Question 7

Direction of line $L$ is $\vec{v} = Q - P = (1, 1, 2)$. Equation of $L$: $\vec{r} = (1, 2, -1) + \lambda(1, 1, 2)$. Let $S = (1+\lambda, 2+\lambda, -1+2\lambda)$. $\vec{RS} = (\lambda-3, \lambda+3, 2\lambda-6)$. Since $RS \perp L$, $\vec{RS} \cdot \vec{v} = 0 \Rightarrow (\lambda-3) + (\lambda+3) + 2(2\lambda-6) = 0 \Rightarrow 6\lambda = 12 \Rightarrow \lambda = 2$. So $S = (3, 4, 3)$. $S$ divides $PT$ internally in ratio $1:2$: $S = \frac{2P + T}{3} \Rightarrow T = 3S - 2P = 3(3, 4, 3) - 2(1, 2, -1) = (7, 8, 11)$. Area of $\Delta PRT$: Base $PT = |T - P| = \sqrt{6^2 + 6^2 + 12^2} = \sqrt{216} = 6\sqrt{6}$. Altitude $RS = |S - R| = \sqrt{(3-4)^2 + (4+1)^2 + (3-5)^2} = \sqrt{1 + 25 + 4} = \sqrt{30}$. Area $= \frac{1}{2} \times 6\sqrt{6} \times \sqrt{30} = 3\sqrt{180} = 18\sqrt{5}$. (D is correct). Orthocentre $H$ lies on altitude $RS$: $H = R + k(S - R) = (4, -1, 5) + k(-1, 5, -2) = (4-k, -1+5k, 5-2k)$. Altitude from $P$ to $RT$: $\vec{PH} \cdot \vec{RT} = 0$. $\vec{RT} = (3, 9, 6)$. $\vec{PH} = (3-k, -3+5k, 6-2k)$. $(3-k)(3) + (-3+5k)(9) + (6-2k)(6) = 0 \Rightarrow 9 - 3k - 27 + 45k + 36 - 12k = 0 \Rightarrow 30k + 18 = 0 \Rightarrow k = -\frac{3}{5}$. $H = (4 + 3/5, -1 - 3, 5 + 6/5) = (23/5, -4, 31/5)$. (A is correct).