If the point of intersection of the lines $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$ and $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$ lies on the $xy$-plane, then the value of $a+b$ is:

$A$ line with positive direction cosines passes through the point $P(2,1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ equals $\qquad$ units.

The equation of the plane passing through the intersection of the planes $x + 2y + 3z + 4 = 0$ and $4x + 3y + 2z + 1 = 0$ and the origin is

Find the coordinates of the foot of the perpendicular drawn from the point $P(-1, 1, 2)$ to the plane $2x - 3y + z - 11 = 0$.

The angle between the line $\frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 2}{c}$ and the plane $ax + by + cz + 6 = 0$ is ......... $^o$

If the line segment joining the points $P(2, 4, 1)$ and $Q(3, 8, 1)$ is divided by the plane $3x - ky - 6z = 0$ externally in the ratio $4:5$, then $k=$