The vertices B and C of a Delta ABC lie on th | vedclass

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(B) Let the line be $L: \frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4} = \lambda$. Any point on this line is $P(3\lambda - 2, 1, 4\lambda)$.Let $D$ b

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$\sqrt{34}$

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(B) Let the line be $L: \frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4} = \lambda$. Any point on this line is $P(3\lambda - 2, 1, 4\lambda)$.Let $D$ be the foot of the perpendicular from $A(1, -1, 2)$ to the line $L$. Since $D$ lies on $L$,$D = (3\lambda - 2, 1, 4\lambda)$.The direction vector of the line $L$ is $\vec{v} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.The vector $\vec{AD} = (3\lambda - 2 - 1)\hat{i} + (1 - (-1))\hat{j} + (4\lambda - 2)\hat{k} = (3\lambda - 3)\hat{i} + 2\hat{j} + (4\lambda - 2)\hat{k}$.Since $\vec{AD} \perp \vec{v}$,their dot product is zero:$3(3\lambda - 3) + 0(2) + 4(4\lambda - 2) = 0$$9\lambda - 9 + 16\lambda - 8 = 0$$25\lambda = 17 \implies \lambda = \frac{17}{25}$.Substituting $\lambda$ back into $\vec{AD}$:$\vec{AD} = (3(\frac{17}{25}) - 3)\hat{i} + 2\hat{j} + (4(\frac{17}{25}) - 2)\hat{k} = (\frac{51-75}{25})\hat{i} + 2\hat{j} + (\frac{68-50}{25})\hat{k} = -\frac{24}{25}\hat{i} + 2\hat{j} + \frac{18}{25}\hat{k}$.The length of the altitude $AD = |\vec{AD}| = \sqrt{(-\frac{24}{25})^2 + 2^2 + (\frac{18}{25})^2} = \sqrt{\frac{576}{625} + 4 + \frac{324}{625}} = \sqrt{\frac{900}{625} + 4} = \sqrt{\frac{36}{25} + 4} = \sqrt{\frac{36+100}{25}} = \sqrt{\frac{136}{25}} = \frac{\sqrt{4 	imes 34}}{5} = \frac{2\sqrt{34}}{5}$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AD = \frac{1}{2} 	imes 5 	imes \frac{2\sqrt{34}}{5} = \sqrt{34}$.

The vertices $B$ and $C$ of a $\Delta ABC$ lie on the line $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $BC = 5 \text{ units}$. If the vertex $A$ is $(1, -1, 2)$,then the area of this triangle (in $\text{sq. units}$) is:

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