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(D) Let the line be $L: \frac{x}{1} = \frac{y-3}{1} = \frac{z-1}{-1} = t$. Any point on the line is $S(t, t+3, 1-t)$.Vector $\vec{QS} = (t-3, t+6, -t)

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$81$

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(D) Let the line be $L: \frac{x}{1} = \frac{y-3}{1} = \frac{z-1}{-1} = t$. Any point on the line is $S(t, t+3, 1-t)$.Vector $\vec{QS} = (t-3, t+6, -t)$. The direction vector of the line is $\vec{v} = (1, 1, -1)$.Since $\vec{QS} \perp \vec{v}$,we have $(t-3)(1) + (t+6)(1) + (-t)(-1) = 0$,which gives $t-3+t+6+t = 0 \implies 3t+3=0 \implies t=-1$.Thus,the foot of the perpendicular is $S(-1, 2, 2)$.Since $S$ is the midpoint of $PQ$,we have $\frac{\alpha+3}{2} = -1, \frac{\beta-3}{2} = 2, \frac{\gamma+1}{2} = 2$,so $P(-5, 7, 3)$.Vector $\vec{RQ} = (3-2, -3-5, 1-(-1)) = (1, -8, 2)$ and $\vec{RP} = (-5-2, 7-5, 3-(-1)) = (-7, 2, 4)$.The area of $	riangle PQR = \frac{1}{2} |\vec{RQ} 	imes \vec{RP}|$.$\vec{RQ} 	imes \vec{RP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -8 & 2 \ -7 & 2 & 4 \end{vmatrix} = \hat{i}(-32-4) - \hat{j}(4+14) + \hat{k}(2-56) = -36\hat{i} - 18\hat{j} - 54\hat{k}$.Area $\lambda = \frac{1}{2} \sqrt{(-36)^2 + (-18)^2 + (-54)^2} = \frac{1}{2} \sqrt{1296 + 324 + 2916} = \frac{1}{2} \sqrt{4536} = \frac{1}{2} \sqrt{324 	imes 14} = \frac{18}{2} \sqrt{14} = 9\sqrt{14}$.Given $\lambda^2 = 14K$,we have $(9\sqrt{14})^2 = 81 	imes 14 = 14K$,so $K = 81$.

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line $\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}$ and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$,then $K$ is equal to:

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