MCA Entrance Previous Year Question Bank

If the equation of the plane passing through the line of intersection of the planes $2x-y+z=3$ and $4x-3y+5z+9=0$ and parallel to the line $\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$ is $ax+by+cz+6=0$, then $a+b+c$ is equal to:

Let $O$ be the origin, and $M$ and $N$ be the points on the lines $$\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$$ and $$\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$$ respectively such that $MN$ is the shortest distance between the given lines. Then $\overrightarrow{OM}\cdot \overrightarrow{ON}$ is equal to

$\text{Let }{L}_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ $\text{and }{L}_2:\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\text{ be two lines.}$ $\text{Then which of the following points lies on the line of the shortest distance}$ $\text{between }{L}_1\text{ and }{L}_2?$