Let the line frac{x}{1}=frac{6-y}{2}=frac{z+8 | vedclass

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(A) The given lines are:$L_1: \frac{x}{1} = \frac{y-6}{-2} = \frac{z+8}{5} = \lambda$$L_2: \frac{x-5}{4} = \frac{y-7}{3} = \frac{z+2}{1} = \mu$$L_3: \

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$4$

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(A) The given lines are:$L_1: \frac{x}{1} = \frac{y-6}{-2} = \frac{z+8}{5} = \lambda$$L_2: \frac{x-5}{4} = \frac{y-7}{3} = \frac{z+2}{1} = \mu$$L_3: \frac{x+3}{6} = \frac{y-3}{-3} = \frac{z-6}{1} = \gamma$For intersection point $A$ of $L_1$ and $L_2$:$(\lambda, -2\lambda+6, 5\lambda-8) = (4\mu+5, 3\mu+7, \mu-2)$Solving these,we get $\lambda=1$ and $\mu=-1$. Thus,$A = (1, 4, -3)$.For intersection point $B$ of $L_1$ and $L_3$:$(\lambda, -2\lambda+6, 5\lambda-8) = (6\gamma-3, -3\gamma+3, \gamma+6)$Solving these,we get $\lambda=3$ and $\gamma=1$. Thus,$B = (3, 0, 7)$.The mid-point $M$ of $AB$ is $(\frac{1+3}{2}, \frac{4+0}{2}, \frac{-3+7}{2}) = (2, 2, 2)$.The distance of $M(2, 2, 2)$ from the plane $2x-2y+z-14=0$ is given by:$d = \frac{|2(2) - 2(2) + 1(2) - 14|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|4 - 4 + 2 - 14|}{\sqrt{4+4+1}} = \frac{|-12|}{3} = 4$.

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $AB$ from the plane $2x-2y+z=14$ is

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