Consider the lines L_1: x-1=y-2=z and L_2: x- | vedclass

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(B) The line $L_1$ is given by $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-0}{1}$. Let $Q = (\lambda+1, \lambda+2, \lambda)$.Since $PQ \perp L_1$,the vec

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$147$

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(B) The line $L_1$ is given by $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-0}{1}$. Let $Q = (\lambda+1, \lambda+2, \lambda)$.Since $PQ \perp L_1$,the vector $\vec{PQ} = (\lambda+1-5, \lambda+2-1, \lambda-(-3)) = (\lambda-4, \lambda+1, \lambda+3)$ is perpendicular to the direction vector $\vec{v_1} = (1, 1, 1)$.Thus,$(\lambda-4)(1) + (\lambda+1)(1) + (\lambda+3)(1) = 0 \Rightarrow 3\lambda = 0 \Rightarrow \lambda = 0$.So,$Q = (1, 2, 0)$ and $\vec{PQ} = (-4, 1, 3)$.The line $L_2$ is given by $\frac{x-2}{1} = \frac{y-0}{1} = \frac{z-1}{1}$. Let $R = (\mu+2, \mu, \mu+1)$.Since $PR \perp L_2$,the vector $\vec{PR} = (\mu+2-5, \mu-1, \mu+1-(-3)) = (\mu-3, \mu-1, \mu+4)$ is perpendicular to the direction vector $\vec{v_2} = (1, 1, 1)$.Thus,$(\mu-3)(1) + (\mu-1)(1) + (\mu+4)(1) = 0 \Rightarrow 3\mu = 0 \Rightarrow \mu = 0$.So,$R = (2, 0, 1)$ and $\vec{PR} = (-3, -1, 4)$.The area of $	riangle PQR$ is $A = \frac{1}{2} |\vec{PQ} 	imes \vec{PR}|$.$\vec{PQ} 	imes \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -4 & 1 & 3 \ -3 & -1 & 4 \end{vmatrix} = \hat{i}(4 - (-3)) - \hat{j}(-16 - (-9)) + \hat{k}(4 - (-3)) = 7\hat{i} + 7\hat{j} - 7\hat{k}$.$|\vec{PQ} 	imes \vec{PR}| = \sqrt{7^2 + 7^2 + (-7)^2} = \sqrt{49 	imes 3} = 7\sqrt{3}$.$A = \frac{1}{2} 	imes 7\sqrt{3} = \frac{7\sqrt{3}}{2}$.$4A^2 = 4 	imes \frac{49 	imes 3}{4} = 147$.

Consider the lines $L_1: x-1=y-2=z$ and $L_2: x-2=y=z-1$. Let the feet of the perpendiculars from the point $P(5,1,-3)$ on the lines $L_1$ and $L_2$ be $Q$ and $R$ respectively. If the area of the triangle $PQR$ is $A$,then $4A^2$ is equal to:

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