Class 12 Mathematics Practice Paper Set 3 with Solutions

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Class 12 Mathematics · Practice Paper SET-3 with Solutions

CBSE PRACTICE PAPER 2026–27 · MATHEMATICS (SET – 3)

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CLASS : XIISUBJECT : MATHEMATICSROLL NO. : ..................

Time : 3 Hrs.Session : 2026–27 · Full Syllabus · High LevelM.M. : 80

General Instructions

1. This question paper contains 38 questions. All questions are compulsory.
2. The paper is divided into five sections – A, B, C, D and E. The paper is of higher difficulty level (HOTS / competitive pattern).
3. Section A: Q. 1 to 18 are MCQs and Q. 19–20 are Assertion–Reason questions, 1 mark each.
4. Section B: Q. 21 to 25 are short answer questions of 2 marks each.
5. Section C: Q. 26 to 31 are long-type short answer questions of 3 marks each.
6. Section D: Q. 32 to 35 are long answer questions of 5 marks each.
7. Section E: Q. 36 to 38 are case-based questions of 4 marks each, with sub-parts.
8. There is no overall choice. Internal choice is provided in some questions of Sections B, C, D and E.
9. Use of calculator is not permitted. Draw neat figures wherever required.
10. The detailed Answer Key & Solutions are given at the end of this paper.

SECTION A (20 × 1 = 20 Marks)

Q. 1 to Q. 18 : Multiple Choice Questions (choose the correct option)

1. The number of relations on the set {1, 2, 3} that are both reflexive and symmetric is

(a) 4(b) 16(c) 8(d) 64

2. The function f : R → R defined by f(x) = x³ − 3x is

(a) one-one and onto(b) one-one but not onto(c) neither one-one nor onto(d) onto but not one-one

3. The value of tan⁻¹1 + tan⁻¹2 + tan⁻¹3 is

(a) π/2(b) π(c) 3π/4(d) 2π

4. The sum of all solutions of the equation sin⁻¹x + sin⁻¹(1 − x) = cos⁻¹x is

(a) 1/2(b) 1(c) 0(d) 3/2

5. If A is a square matrix such that A² = A, then (I + A)³ − 7A is equal to

(a) A(b) O(c) I(d) 3I

6. If A =

3	−4
1	−1

, then the element in the first row and second column of A²⁵ is

(a) −100(b) 100(c) −50(d) 25

7. If the area of the triangle with vertices (2, −6), (5, 4) and (k, 4) is 35 square units, then the sum of all possible values of k is

(a) 12(b) 10(c) −2(d) 14

8. Let f(x) = x² sin(1/x) for x ≠ 0 and f(0) = 0. Then at x = 0, the function f is

(a) not continuous(b) continuous but not differentiable(c) differentiable and f′(0) = 0(d) differentiable and f′(0) = 1

9. If y = log √[(1 − cos x)/(1 + cos x)], then dy/dx is

(a) sec x(b) −cosec x(c) cot x(d) cosec x

10. The absolute maximum value of f(x) = x − x³ on the interval [0, 1] is

(a) 2√3/9(b) 1/3(c) 2/9(d) 0

11. ∫ cos 2x − cos 2αcos x − cos α dx =

(a) sin x + x cos α + C(b) 2(sin x + x cos α) + C(c) 2(sin x − x cos α) + C(d) 2(cos x + x sin α) + C

12. ∫₀² [x²] dx, where [ · ] denotes the greatest integer function, equals

(a) 5 − √3 + √2(b) 4 − √2 − √3(c) 6 − √2 − √3(d) 5 − √2 − √3

13. The integrating factor of the differential equation (x log x) dy/dx + y = 2 log x is

(a) log x(b) x(c) 1/x(d) x log x

14. If a × b = c × d and a × c = b × d, then

(a) (a − d) is perpendicular to (b − c)(b) a − d = b − c(c) (a − d) is parallel to (b − c)(d) (a − d) is parallel to (b + c)

15. The vectors 2i − j + k, i + 2j − 3k and 3i + λj + 5k are coplanar if λ equals

(a) 4(b) −4(c) −2(d) 2

16. If a line makes angles α, β, γ with the coordinate axes, then sin²α + sin²β + sin²γ is equal to

(a) 1(b) 3(c) 3/2(d) 2

17. If P(A) = 0.6, P(B) = 0.3 and P(A ∩ B) = 0.2, then P(A′ | B′) is

(a) 2/7(b) 3/10(c) 3/7(d) 4/7

18. The mean and variance of a binomial distribution are 4 and 3 respectively. The number of trials n is

(a) 16(b) 12(c) 8(d) 24

Q. 19 and Q. 20 : Assertion–Reason Questions

Options: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

19. Assertion (A): tan⁻¹x + cot⁻¹x = π/2 for every real number x.
Reason (R): The range of the principal value branch of cot⁻¹x is (0, π).

20. Assertion (A): ∫_−a^a f(x) dx = 0 for every continuous function f.
Reason (R): If f is an even function, then ∫_−a^a f(x) dx = 2∫₀^a f(x) dx.

SECTION B (5 × 2 = 10 Marks) — Short Answer Questions

21. Prove that tan⁻¹(1/4) + tan⁻¹(2/9) = ½ tan⁻¹(4/3).(2)

OR

Simplify : tan⁻¹cos x1 + sin x, −π/2 < x < π/2.

22. Differentiate tan⁻¹3x − x³1 − 3x² with respect to x, where |x| < 1/√3.(2)

OR

If x^y = e^{x − y}, show that dy/dx = log x(1 + log x)².

23. Find the intervals in which f(x) = sin x + cos x, 0 ≤ x ≤ 2π, is increasing and decreasing.(2)

OR

The side of an equilateral triangle is increasing at the rate of 3 cm/s. Find the rate at which its area is increasing when the side is 10 cm.

24. Evaluate ∫ (log x)² dx.(2)

OR

Evaluate ∫ dx√(5 − 4x − x²).

25. Find the value of k for which the lines 1 − x3 = y − 22k = z − 32 and x − 13k = y − 11 = 6 − z5 are perpendicular to each other.(2)

OR

Find the angle between the line x − 12 = y + 11 = z − 3−2 and the plane 2x + 3y + 6z = 11.

SECTION C (6 × 3 = 18 Marks) — Long-Type Short Answer Questions

26. Evaluate ∫ x² + 1x⁴ + 1 dx.(3)

27. Evaluate ∫₀^π/2 sin x cos xsin⁴x + cos⁴x dx.(3)

OR

Evaluate ∫_−π/2^π/2 sin²x1 + e^x dx.

28. Solve the differential equation x dy/dx = y (log y − log x + 1), where x, y > 0.(3)

29. Two numbers are selected at random (without replacement) from the first five positive integers 1, 2, 3, 4, 5. Let X denote the larger of the two numbers obtained. Find the probability distribution of X and hence its mean E(X).(3)

30. Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.(3)

31. If A =

2	−1	1
−1	2	−1
1	−1	2

, show that A³ − 6A² + 9A − 4I = O. Hence find A⁻¹.(3)

SECTION D (4 × 5 = 20 Marks) — Long Answer Questions

32. Using integration, find the area of the smaller region bounded by the ellipse x²/4 + y²/9 = 1 and the line x/2 + y/3 = 1.(5)

OR

Using integration, find the area of the region bounded by the parabola y² = 8x and its latus rectum.

33. Determine the values of λ and μ for which the system of equations
x + y + z = 6,   x + 2y + 3z = 10,   x + 2y + λz = μ
has (i) a unique solution, (ii) no solution, (iii) infinitely many solutions. Also solve the system for λ = 4, μ = 12.(5)

OR

Show that the function f : N → Y defined by f(x) = 4x + 3, where Y = {y ∈ N : y = 4x + 3 for some x ∈ N}, is invertible. Find the inverse of f and hence find f⁻¹(19).

34. Show that the semi-vertical angle of the right circular cone of maximum volume and given slant height l is tan⁻¹√2.(5)

OR

Find the point on the curve y² = 2x which is nearest to the point (1, 4).

35. Solve the following linear programming problem graphically: Maximise and minimise Z = 5x + 7y subject to 2x + y ≤ 10, x + 3y ≤ 15, x + y ≥ 4, x ≥ 0, y ≥ 0.(5)

OR

Solve graphically: Minimise Z = 3x + 5y subject to x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0.

SECTION E (3 × 4 = 12 Marks) — Case-Based Questions

36. Case Study – Conical sand pile. Sand is poured on a level floor at the rate of 12 cm³/s. The sand forms a conical pile whose height is always equal to the radius r of its base.

(i) Express the volume V of the pile in terms of r. (1)
(ii) Find the rate at which the radius is increasing when r = 4 cm. (1)
(iii) Find the rate at which the curved surface area of the pile is increasing when r = 4 cm. (2)

OR

(iii) Find the rate at which the area of the base of the pile is increasing when r = 4 cm. (2)

37. Case Study – Guessing in a test. A student answers 4 multiple-choice questions by pure guessing. Each question has 4 options of which exactly one is correct. Let X denote the number of correct answers.

(i) Find P(X = 0). (1)
(ii) Find P(X ≥ 3). (1)
(iii) Find the mean and the variance of X. (2)

OR

(iii) Find the probability that the student gets at least two answers correct. (2)

38. Case Study – Drone and landing pad. A drone hovers at the point P(5, 2, 0) above a flat landing pad which lies in the plane 2x + y − 2z = 3.

(i) Write the direction ratios of the normal to the landing pad. (1)
(ii) Find the perpendicular distance of P from the plane. (1)
(iii) Find the coordinates of the foot of the perpendicular drawn from P to the plane. (2)

OR

(iii) Find the coordinates of the image of P in the plane. (2)

ANSWER KEY & SOLUTIONS · MATHEMATICS PRACTICE PAPER (SET – 3)

www.careerinformationportal.in

CLASS : XIISUBJECT : MATHEMATICSROLL NO. : ..................

Time : 3 Hrs.Class XII · Session : 2026–27M.M. : 80

SECTION A — Answer Key (1 mark each)

Q.1	(c)	Q.2	(d)	Q.3	(b)	Q.4	(a)
Q.5	(c)	Q.6	(a)	Q.7	(b)	Q.8	(c)
Q.9	(d)	Q.10	(a)	Q.11	(b)	Q.12	(d)
Q.13	(a)	Q.14	(c)	Q.15	(b)	Q.16	(d)
Q.17	(c)	Q.18	(a)	Q.19	(b)	Q.20	(d)

Hints for Q.1–20. (1) Diagonal pairs are fixed; each of the 3 unordered pairs {1,2}, {1,3}, {2,3} is either in or out (both orderings together) ⇒ 2³ = 8. (2) f′(x) = 3(x² − 1) changes sign, so f is not one-one (f(0) = f(√3) = 0); a cubic takes every real value, so f is onto. (3) tan⁻¹1 + tan⁻¹2 = π + tan⁻¹[3/(1 − 2)] = π − tan⁻¹3, so the sum is π. (4) x = 0 gives 0 + π/2 = π/2 ✓; x = ½ gives π/6 + π/6 = π/3 = cos⁻¹½ ✓ ⇒ sum = ½. (5) (I + A)³ = I + 3A + 3A² + A³ = I + 7A (as A² = A³ = A) ⇒ result I. (6) A = I + N with N = [(2, −4), (1, −2)], N² = O, so Aⁿ = I + nN; the (1, 2) element is −4n = −100. (7) ½|2(4 − 4) + 5(4 + 6) + k(−6 − 4)| = 35 ⇒ |50 − 10k| = 70 ⇒ k = −2 or 12 ⇒ sum = 10. (8) |f(x)/x| ≤ |x| → 0, so f′(0) = lim x sin(1/x) = 0. (9) The expression is log tan(x/2); its derivative = ½sec²(x/2)/tan(x/2) = 1/sin x = cosec x. (10) f′ = 1 − 3x² = 0 at x = 1/√3; f = 1/√3 − 1/(3√3) = 2√3/9; the end values are 0. (11) cos 2x − cos 2α = 2(cos x − cos α)(cos x + cos α) ⇒ ∫2(cos x + cos α)dx. (12) [x²] = 0, 1, 2, 3 on [0,1), [1,√2), [√2,√3), [√3,2) ⇒ (√2 − 1) + 2(√3 − √2) + 3(2 − √3) = 5 − √2 − √3. (13) IF = e^{∫dx/(x log x)} = e^{log log x} = log x. (14) (a − d) × (b − c) = (a×b − c×d) − (a×c − b×d) = 0. (15) Determinant = 28 + 7λ = 0 ⇒ λ = −4. (16) Σsin² = 3 − Σcos² = 3 − 1 = 2. (17) P(A ∪ B) = 0.7, P(A′ ∩ B′) = 0.3, P(B′) = 0.7 ⇒ 3/7. (18) np = 4, npq = 3 ⇒ q = 3/4, p = 1/4, n = 16. (19) (b): the identity holds ∀ x ∈ R (A true) and R is true, but R is not the reason for A. (20) (d): e.g. f(x) = x² gives 2a³/3 ≠ 0, so A is false; R is true.

SECTION B — Solutions (2 marks each)

21. tan⁻¹(1/4) + tan⁻¹(2/9) = tan⁻¹[(1/4 + 2/9)/(1 − 2/36)] = tan⁻¹[(17/36)/(34/36)] = tan⁻¹(½) (1). Also 2 tan⁻¹(½) = tan⁻¹[1/(1 − ¼)] = tan⁻¹(4/3), so tan⁻¹(½) = ½ tan⁻¹(4/3). Hence proved (1). OR: cos x/(1 + sin x) = [cos²(x/2) − sin²(x/2)]/[cos(x/2) + sin(x/2)]² = [cos(x/2) − sin(x/2)]/[cos(x/2) + sin(x/2)] = tan(π/4 − x/2) (1). Since −π/2 < x < π/2, (π/4 − x/2) ∈ (0, π/2), so the value is π/4 − x/2 (1).

22. Put x = tan θ, θ = tan⁻¹x ∈ (−π/6, π/6). Then (3x − x³)/(1 − 3x²) = tan 3θ (1), so y = tan⁻¹(tan 3θ) = 3θ = 3 tan⁻¹x and dy/dx = 3/(1 + x²) (1). OR: y log x = x − y ⇒ y(1 + log x) = x ⇒ y = x/(1 + log x) (1). dy/dx = [(1 + log x) − x(1/x)]/(1 + log x)² = log x/(1 + log x)² (1).

23. f′(x) = cos x − sin x = 0 ⇒ tan x = 1 ⇒ x = π/4, 5π/4 (½). Sign of f′: positive on [0, π/4) and (5π/4, 2π]; negative on (π/4, 5π/4) (1). So f is increasing on [0, π/4] ∪ [5π/4, 2π] and decreasing on [π/4, 5π/4] (½). OR: A = (√3/4)a² ⇒ dA/dt = (√3/2)a · da/dt (1) = (√3/2)(10)(3) = 15√3 cm²/s (1).

24. Integrating by parts (taking 1 as the second function): ∫(log x)²dx = x(log x)² − ∫2 log x dx (1) = x(log x)² − 2(x log x − x) + C = x(log x)² − 2x log x + 2x + C (1). OR: 5 − 4x − x² = 9 − (x + 2)² (1) ⇒ ∫ = sin⁻¹((x + 2)/3) + C (1).

25. Direction ratios: first line (−3, 2k, 2) [since (1 − x)/3 = (x − 1)/(−3)]; second line (3k, 1, −5) (½). Perpendicular: (−3)(3k) + (2k)(1) + (2)(−5) = 0 (1) ⇒ −7k = 10 ⇒ k = −10/7 (½). OR: b = (2, 1, −2), n = (2, 3, 6); |b| = 3, |n| = 7, b·n = 4 + 3 − 12 = −5 (1). sin θ = 5/21 ⇒ θ = sin⁻¹(5/21) (1).

SECTION C — Solutions (3 marks each)

26. Divide numerator and denominator by x²: I = ∫(1 + 1/x²)/(x² + 1/x²)dx (1). Put t = x − 1/x, dt = (1 + 1/x²)dx, and x² + 1/x² = t² + 2 (1). I = ∫dt/(t² + 2) = (1/√2) tan⁻¹[(x² − 1)/(√2 x)] + C (1).

27. Put sin²x = t, dt = 2 sin x cos x dx; sin⁴x + cos⁴x = t² + (1 − t)² = 2t² − 2t + 1 (1). I = ½∫₀¹ dt/(2t² − 2t + 1) = ¼∫₀¹ dt/[(t − ½)² + ¼] = ¼ · 2[tan⁻¹(2t − 1)]₀¹ (1) = ½(π/4 + π/4) = π/4 (1).

OR

I = ∫_−π/2^π/2 sin²x/(1 + e^x)dx. Replacing x by −x: I = ∫ sin²x · e^x/(1 + e^x)dx (1). Adding: 2I = ∫_−π/2^π/2 sin²x dx = 2∫₀^π/2 sin²x dx = π/2 (1). So I = π/4 (1).

28. dy/dx = (y/x)(log(y/x) + 1): a homogeneous equation. Put y = vx: v + x dv/dx = v(log v + 1) ⇒ x dv/dx = v log v (1). Separating: dv/(v log v) = dx/x ⇒ log|log v| = log x + log C (1) ⇒ log v = Cx ⇒ y = x e^Cx (1).

29. Total outcomes = C(5, 2) = 10. X can be 2, 3, 4, 5 (½). P(X = 2) = 1/10 (pair {1,2}); P(X = 3) = 2/10 ({1,3},{2,3}); P(X = 4) = 3/10; P(X = 5) = 4/10 (1½). E(X) = 2(1/10) + 3(2/10) + 4(3/10) + 5(4/10) = (2 + 6 + 12 + 20)/10 = 4 (1).

30. Normals of the given planes: n₁ = (1, 2, 3), n₂ = (3, 3, 1). The required normal is n₁ × n₂ = i(2 − 9) − j(1 − 9) + k(3 − 6) = (−7, 8, −3) (1½). Plane: −7(x + 1) + 8(y − 3) − 3(z − 2) = 0 (1) ⇒ −7x + 8y − 3z − 25 = 0, i.e. 7x − 8y + 3z + 25 = 0 (½).

31. A² =

6	−5	5
−5	6	−5
5	−5	6

, A³ = A²A =

22	−21	21
−21	22	−21
21	−21	22

(1). A³ − 6A² + 9A − 4I: e.g. (1,1): 22 − 36 + 18 − 4 = 0; (1,2): −21 + 30 − 9 = 0; (2,2): 22 − 36 + 18 − 4 = 0, and similarly all entries are 0, so the result is O (1). Multiplying by A⁻¹: A² − 6A + 9I − 4A⁻¹ = O ⇒ A⁻¹ = ¼(A² − 6A + 9I) = 14

3	1	−1
1	3	1
−1	1	3

(1).

SECTION D — Solutions (5 marks each)

32. The line x/2 + y/3 = 1 meets the ellipse at (2, 0) and (0, 3) (1). In the first quadrant, the ellipse is y = 3√(1 − x²/4) = (3/2)√(4 − x²) and the line is y = 3(1 − x/2) (1). Area = ∫₀²[(3/2)√(4 − x²) − 3(1 − x/2)]dx (1). ∫₀²(3/2)√(4 − x²)dx = (3/2)·π = 3π/2 (quarter of a circle of radius 2) (1); ∫₀²3(1 − x/2)dx = 3[x − x²/4]₀² = 3. Area = 3π/2 − 3 = (3/2)(π − 2) square units (1).

OR

y² = 8x ⇒ a = 2; the latus rectum is x = 2, meeting the parabola at (2, ±4) (1). By symmetry, Area = 2∫₀²2√2 √x dx (2) = 4√2 · (2/3)x^3/2|₀² = 4√2 · (2/3) · 2√2 = 32/3 square units (2).

33. The system is AX = B with |A| =

1	1	1
1	2	3
1	2	λ

= 1(2λ − 6) − 1(λ − 3) + 1(0) = λ − 3 (1½). (i) Unique solution: |A| ≠ 0 ⇒ λ ≠ 3 (any μ) (1). For λ = 3: R₃ − R₂ gives 0·z = μ − 10. (ii) No solution: λ = 3, μ ≠ 10 (1). (iii) Infinitely many solutions: λ = 3, μ = 10 (½). For λ = 4, μ = 12: subtracting the 2nd from the 3rd equation gives z = 2; then y + 2z = 4 ⇒ y = 0; x = 6 − 0 − 2 = 4. x = 4, y = 0, z = 2 (1).

OR

One-one: f(x₁) = f(x₂) ⇒ 4x₁ + 3 = 4x₂ + 3 ⇒ x₁ = x₂ (1½). Onto: every y ∈ Y has the form 4x + 3 for some x ∈ N, i.e. y = f(x) (1). So f is bijective and hence invertible. Define g : Y → N by g(y) = (y − 3)/4: g(f(x)) = x and f(g(y)) = y, so f⁻¹(y) = (y − 3)/4 (1½). f⁻¹(19) = 16/4 = 4 (1).

34. Let the semi-vertical angle be α, so r = l sin α and h = l cos α (1). V = ⅓πr²h = (π/3)l³ sin²α cos α (1). dV/dα = (π/3)l³[2 sin α cos²α − sin³α] = (π/3)l³ sin α (2cos²α − sin²α) (1). dV/dα = 0 ⇒ tan²α = 2 ⇒ tan α = √2 (α is acute). For α slightly less than tan⁻¹√2, dV/dα > 0, and for α slightly more, dV/dα < 0, so V is maximum (1½). Hence the semi-vertical angle is tan⁻¹√2 (½).

OR

Let P(y²/2, y) be a point on the curve. D² = (y²/2 − 1)² + (y − 4)² (1). d(D²)/dy = 2(y²/2 − 1)y + 2(y − 4) = y³ − 2y + 2y − 8 = y³ − 8 (1). So y = 2; second derivative 3y² = 12 > 0 ⇒ minimum (1½). The nearest point is (2, 2) (1½).

35. Draw 2x + y = 10 (through (5, 0), (0, 10)), x + 3y = 15 (through (15, 0), (0, 5)) and x + y = 4 (through (4, 0), (0, 4)). The feasible region is the pentagon with vertices A(4, 0), B(5, 0), C(3, 4), D(0, 5), E(0, 4); C is the intersection of 2x + y = 10 and x + 3y = 15 (x = 3, y = 4) (2). Z = 5x + 7y: A = 20, B = 25, C = 15 + 28 = 43, D = 35, E = 28 (2). Maximum Z = 43 at (3, 4); minimum Z = 20 at (4, 0) (1).

OR

Draw x + 3y = 3 and x + y = 2. The feasible region is unbounded with corner points A(3, 0), B(3/2, 1/2) (solving x + 3y = 3 and x + y = 2) and C(0, 2) (2). Z = 3x + 5y: A = 9, B = 9/2 + 5/2 = 7, C = 10 (1). Since the region is unbounded, check 3x + 5y < 7: this open half-plane has no point in common with the feasible region. Hence the minimum value of Z is 7 at (3/2, 1/2) (2).

SECTION E — Solutions (4 marks each)

36. (i) h = r, so V = ⅓πr²h = πr³/3 (1). (ii) dV/dt = πr² dr/dt = 12; at r = 4: dr/dt = 12/(16π) = 3/(4π) cm/s (1). (iii) Slant height l = √(r² + h²) = √2 r, so S = πrl = √2 πr² (1); dS/dt = 2√2 πr dr/dt = 2√2 π(4)(3/(4π)) = 6√2 cm²/s (1). OR (iii): A = πr² ⇒ dA/dt = 2πr dr/dt (1) = 2π(4)(3/(4π)) = 6 cm²/s (1).

37. X is binomial with n = 4, p = 1/4, q = 3/4. (i) P(X = 0) = (3/4)⁴ = 81/256 (1). (ii) P(X ≥ 3) = P(3) + P(4) = 4(1/4)³(3/4) + (1/4)⁴ = 12/256 + 1/256 = 13/256 (1). (iii) Mean = np = 1 (1); variance = npq = 4 · ¼ · ¾ = 3/4 (1). OR (iii): P(X ≥ 2) = 1 − P(0) − P(1) = 1 − 81/256 − 108/256 (1) = 67/256 (1).

38. (i) The normal has direction ratios (2, 1, −2) (1). (ii) d = |2(5) + 2 − 2(0) − 3|/√(4 + 1 + 4) = 9/3 = 3 units (1). (iii) The foot Q = P − λ(2, 1, −2) with λ = (value of plane at P)/|n|² = 9/9 = 1, so Q = (5 − 2, 2 − 1, 0 + 2) = (3, 1, 2) (2) (check: 6 + 1 − 4 = 3 ✓). OR (iii): Image P′ = P − 2λ(2, 1, −2) = (5 − 4, 2 − 2, 0 + 4) = (1, 0, 4) (2).