What is the distance of the point $(2, 3, 4)$ from the plane $3x - 6y + 2z + 11 = 0$?

The equation of the plane passing through the point $(2, 5, -3)$ and perpendicular to the planes $x + 2y + 2z = 1$ and $x - 2y + 3z = 4$ is:

$A$ vector parallel to the line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is

Show that the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar.

From a point $P(\lambda, \lambda, \lambda)$,perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y=x, z=1$ and $y=-x, z=-1$. If $P$ is such that $\angle QPR$ is a right angle,then the possible value$(s)$ of $\lambda$ is(are)

The plane $3x + 4y + 6z + 7 = 0$ is rotated about the line $r = (\hat{i} + 2\hat{j} - 3\hat{k}) + t(2\hat{i} - 3\hat{j} + \hat{k})$ until the plane passes through the origin. The equation of the plane in the new position is