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(D) Let the line be $L: \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2} = \lambda$. Any point on the line is $P(7\lambda+3, -\lambda+2, -2\lambda-1)$.Sinc

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$3 \sqrt{30}$

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(D) Let the line be $L: \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2} = \lambda$. Any point on the line is $P(7\lambda+3, -\lambda+2, -2\lambda-1)$.Since $P$ is the foot of the perpendicular from $Q(10,-3,-1)$ to the line,the vector $\vec{QP}$ must be perpendicular to the direction vector of the line $\vec{v} = 7\hat{i} - \hat{j} - 2\hat{k}$.$\vec{QP} = (7\lambda+3-10)\hat{i} + (-\lambda+2+3)\hat{j} + (-2\lambda-1+1)\hat{k} = (7\lambda-7)\hat{i} + (-\lambda+5)\hat{j} - 2\lambda\hat{k}$.Since $\vec{QP} \cdot \vec{v} = 0$,we have $7(7\lambda-7) - 1(-\lambda+5) - 2(-2\lambda) = 0$.$49\lambda - 49 + \lambda - 5 + 4\lambda = 0 \Rightarrow 54\lambda - 54 = 0 \Rightarrow \lambda = 1$.Thus,$P = (7(1)+3, -1+2, -2(1)-1) = (10, 1, -3)$.Now,$\vec{PQ} = (10-10)\hat{i} + (-3-1)\hat{j} + (-1-(-3))\hat{k} = -4\hat{j} + 2\hat{k}$.And $\vec{PR} = (3-10)\hat{i} + (-2-1)\hat{j} + (1-(-3))\hat{k} = -7\hat{i} - 3\hat{j} + 4\hat{k}$.The area of $	riangle PQR = \frac{1}{2} |\vec{PQ} 	imes \vec{PR}|$.$\vec{PQ} 	imes \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -4 & 2 \ -7 & -3 & 4 \end{vmatrix} = \hat{i}(-16 - (-6)) - \hat{j}(0 - (-14)) + \hat{k}(0 - 28) = -10\hat{i} - 14\hat{j} - 28\hat{k}$.Magnitude $|\vec{PQ} 	imes \vec{PR}| = \sqrt{(-10)^2 + (-14)^2 + (-28)^2} = \sqrt{100 + 196 + 784} = \sqrt{1080} = \sqrt{36 	imes 30} = 6\sqrt{30}$.Area $= \frac{1}{2} 	imes 6\sqrt{30} = 3\sqrt{30}$.

Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right-angled triangle $PQR$,where $R$ is the point $(3,-2,1)$,is

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