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(B) Let $M$ be the foot of the perpendicular from $B(1,2,3)$ to the line $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} = \lambda$.Any point on the li

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$21$

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(B) Let $M$ be the foot of the perpendicular from $B(1,2,3)$ to the line $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} = \lambda$.Any point on the line is $M(3\lambda+6, 2\lambda+7, -2\lambda+7)$.The vector $\overrightarrow{BM} = (3\lambda+6-1)\hat{i} + (2\lambda+7-2)\hat{j} + (-2\lambda+7-3)\hat{k} = (3\lambda+5)\hat{i} + (2\lambda+5)\hat{j} + (-2\lambda+4)\hat{k}$.Since $\overrightarrow{BM}$ is perpendicular to the direction vector of the line $\vec{v} = 3\hat{i} + 2\hat{j} - 2\hat{k}$,we have $\overrightarrow{BM} \cdot \vec{v} = 0$.$3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0$.$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \implies 17\lambda + 17 = 0 \implies \lambda = -1$.Substituting $\lambda = -1$,we get $\overrightarrow{BM} = (3(-1)+5)\hat{i} + (2(-1)+5)\hat{j} + (-2(-1)+4)\hat{k} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.The length of the altitude $BM = |\overrightarrow{BM}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.The area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AC 	imes BM = \frac{1}{2} 	imes 6 	imes 7 = 21$ sq. units.

Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is

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