Question 5

**Direction of L₁:** cross product of normals (2,3,1) and (1,2,1): (3−2, 1−2, 4−3) = (1, −1, 1). **Line L₂:** (2+t, −1−t, 3+t). Substituting into 2x + y − 2z = 6: 2(2+t) + (−1−t) − 2(3+t) = −3 − t = 6 ⇒ t = −9. So Q = (−7, 8, −6), and PQ = 9·√3. **(A) true.** **Foot of perpendicular R:** signed distance of P from M = (4 − 1 − 6 − 6)/3 = −3, so PR = 3 and R = (2,−1,3) + 3·(2,1,−2)/3 = (4, 0, 1). (Check: 8 + 0 − 2 = 6 ✓) **QR:** Since PR ⊥ plane, triangle PQR is right-angled at R: QR = √(PQ² − PR²) = √(243 − 9) = √234. **(B) false** (not 15). **Area:** ½ · PR · QR = ½ · 3 · √234 = (3/2)√234. **(C) true.** **Angle between PQ and PR:** cos θ = PR/PQ = 3/(9√3) = 1/(3√3), not 1/(2√3). **(D) false.**

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