If the line $\frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + \lambda}{-2}$ lies in the plane $2x - 4y + 3z = 2$,then the shortest distance between this line and the line $\frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$ is

The foot of the perpendicular drawn from $A(1, 2, 2)$ onto the plane $x+2y+2z-5=0$ is $B(\alpha, \beta, \gamma)$. If $\pi(x, y, z) \equiv x+2y+2z+5=0$ is a plane,then $-\pi(A) : \pi(B) =$ ?

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{-5}$ in the plane $2x - y + z + 3 = 0$ is the line:

Let $O(\overrightarrow{0}), A(\hat{i}+2 \hat{j}+\hat{k}), B(-2 \hat{i}+3 \hat{k}), C(-2 \hat{i}+\hat{j}), D(4 \hat{k})$ be the position vectors of the points $O, A, B, C$ and $D$. If a line passing through $A$ and $B$ intersects the plane passing through $O, C$ and $D$ at the point $R$,then the position vector of $R$ is:

The coordinates of the point,where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}$ meets the plane $2x+4y-z=3$,are