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

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(A) The line equation is $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda$. Any point on this line is $P(\lambda, 2\lambda+1, 3\lambda+2)$.Since $B(

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$5\sqrt{13}$

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(A) The line equation is $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda$. Any point on this line is $P(\lambda, 2\lambda+1, 3\lambda+2)$.Since $B(4, 9, \alpha)$ lies on the line,we have $\frac{4}{1} = \frac{9-1}{2} = \frac{\alpha-2}{3} \Rightarrow 4 = 4 = \frac{\alpha-2}{3} \Rightarrow \alpha = 14$.Let $AD$ be the altitude from $A(1, 6, 3)$ to the line $BC$. $D$ is the projection of $A$ on the line,so $D(\lambda, 2\lambda+1, 3\lambda+2)$.The vector $\vec{AD} = (\lambda-1)\hat{i} + (2\lambda+1-6)\hat{j} + (3\lambda+2-3)\hat{k} = (\lambda-1)\hat{i} + (2\lambda-5)\hat{j} + (3\lambda-1)\hat{k}$.Since $\vec{AD}$ is perpendicular to the line direction vector $\vec{v} = \hat{i} + 2\hat{j} + 3\hat{k}$,their dot product is zero:$(\lambda-1)(1) + (2\lambda-5)(2) + (3\lambda-1)(3) = 0$$\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0 \Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$.Thus,$D = (1, 2(1)+1, 3(1)+2) = (1, 3, 5)$.The length of altitude $AD = \sqrt{(1-1)^2 + (3-6)^2 + (5-3)^2} = \sqrt{0 + 9 + 4} = \sqrt{13}$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AD = \frac{1}{2} 	imes 10 	imes \sqrt{13} = 5\sqrt{13}$ sq. units.

The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$. The coordinates of $A$ and $B$ are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively,and $C$ is at a distance of $10$ units from $B$. The area (in sq. units) of $\Delta ABC$ is:

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