Let the foot of the perpendicular from the point $P (3, -2, -9)$ on the plane passing through the points $A (-1, -2, -3)$,$B (9, 3, 4)$,and $C (9, -2, 1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is:

$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ equals

The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $3x+4y+5z=2$ and perpendicular to the $XY$-plane is

Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})$ and the plane $\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$.

$A$ 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

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 $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is: