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

A plane $\pi_1$ passing through the point $3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ is perpendicular to the vector $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and another plane $\pi_2$ passing through the point $2 \hat{\mathbf{i}}+7 \hat{\mathbf{k}}-8 \hat{\mathbf{k}}$ is perpendicular to the vector $3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$. If $p_1$ and $p_2$ are the perpendicular distances from the origin to the planes $\pi_1$ and $\pi_2$ respectively, then $p_1-p_2=$

$A(2,3, k), B(-1, k,-1)$ and $C(4,-3,2)$ are the vertices of $\triangle A B C$. If $A B=A C$ and $k>0$, then $\triangle A B C$ is

If $a, b$ and $c$ are the intercepts made on $X, Y, Z$-axes respectively by the plane passing through the points $(1,0,-2),(3,-1,2)$ and $(0,-3,4)$, then $3 a+4 b+7 c=$