The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .

Find the distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$.

If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$ $(a > 0)$ makes an angle $30^{\circ}$ with the plane $y - z + 2 = 0$,then the direction cosines of the line are:

$A$ line $L$ is passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-3 \hat{k}$ and is parallel to the vector $2 \hat{i}+\hat{j}+2 \hat{k}$. $A$ plane $\pi$ is passing through the points $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}-\hat{k}$ and is parallel to the vector $\hat{i}-2 \hat{j}$. Then the point where this plane $\pi$ meets the line $L$ is

If the equation of a plane $P,$ passing through the intersection of the planes $x+4y-z+7=0$ and $3x+y+5z=8$ is $ax+by+6z=15$ for some $a, b \in R,$ then the distance of the point $(3,2,-1)$ from the plane $P$ 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 $-\pi(A) : \pi(B) =$ ?