Find the distance of the point $-\hat{i} - 5\hat{j} - 10\hat{k}$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5$.

If $P=(0,1,0)$ and $Q=(0,0,1)$,then the length of the projection of the line segment $PQ$ on the plane $x+y+z=3$ is:

For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to:

Let the lines $l_1: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ and $l_2: 3x+2y+z-2=0=x-3y+2z-13$ be coplanar. If the point $P(a, b, c)$ on $l_1$ is nearest to the point $Q(-4, -3, 2)$,then $|a|+|b|+|c|$ is equal to

If the line passing through the points $(5, 1, a)$ and $(3, b, 1)$ intersects the plane at the point $(0, \frac{17}{2}, -\frac{13}{2})$,then find the values of $a$ and $b$.

The distance of the point $(1, -2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2z = 3$ and $2x - 2y + z + 12 = 0$ is: