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 $\mathrm{A}$ and $\mathrm{B}$ respectively. Then the distance of the mid-point of the line segment $\mathrm{AB}$ from the plane $2 x-2 y+z=14$ is :

We have, $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the line <br/><br/>$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ <br/><br/>$$ \begin{array}{ll} & \text { Now, } \frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}=\lambda ...........(i)\\\\ & \Rightarrow x=\lambda, y=6-2 \lambda, z=5 \lambda-8 \end{array} $$ <br/><br/>$$ \begin{array}{ll} &\text { Also, } \frac{x-5}{4} =\frac{y-7}{3}=\frac{z+2}{1}=k ...........(ii)\\\\ &\Rightarrow x =4 k+5, y=3 k+7, z=k-2 \end{array} $$ <br/><br/>$$ \begin{array}{rlrl} & \frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}=\mu ..........(iii)\\\\ &\Rightarrow x = 6 \mu-3, y=3-3 \mu, z=\mu+6 \end{array} $$ <br/><br/>On solving Eqs. (i) and (ii), we get $\lambda=1, k=-1$ <br/><br/>$\therefore$ Co-ordinate of $A$ is $(1,4,-3)$ <br/><br/>On solving Eqs. (i) and (iii), we get $\lambda=3, \mu=1$ <br/><br/>$\therefore$ Co-ordinate of $\beta$ is $(3,0,7)$ <br/><br/>Co-ordinate of mid-point of $A B$ is $\left(\frac{1+3}{2}, \frac{4+0}{2}, \frac{-3+7}{2}\right)$ or $(2,2,2)$ <br/><br/>Perpendicular distance of mid-point of $A B$ from the plane $2 x-2 y+z=14$ is <br/><br/>$\frac{|2(2)-2(2)+2-14|}{\sqrt{2^2+(-2)^2+1^2}}=4$