$A$ point $P$ lies on a line passing through $Q(1, -2, 3)$ and is parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$. If $P$ lies on the plane $2x + 3y - 4z + 22 = 0$, then the length of the segment $PQ$ is:

The ratio in which the line segment joining the points $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is:

The plane through the intersection of planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y$-axis also passes through the point

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, z=1$,then $\alpha+\beta=$

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

Find the distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$.