The equation of the plane containing the line of intersection of the planes $2x - y = 0$ and $y - 3z = 0$ and perpendicular to the plane $4x + 5y - 3z - 8 = 0$ is

Let $\Pi$ be a plane containing the points $(0,-5,-1), (1,-2,5), (-3,5,0)$ and $L$ be a line passing through the point $(0,-5,-1)$ and parallel to the vector $\hat{i}+5\hat{j}-6\hat{k}$. Then the length of the projection of the unit normal vector to the plane $\Pi$ on the line $L$ is

The vector equation of the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $\overrightarrow{r} \cdot (\hat{i} - 2\hat{j}) = -2$,and the point $(1, 0, 2)$ is:

Find the length and the foot of the perpendicular from the point $\left(1, \frac{3}{2}, 2\right)$ to the plane $2x - 2y + 4z + 5 = 0$.

$A$ line $l$ passes through the origin and is perpendicular to the lines $l_1 = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$ and $l_2 = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$.
Statement $1$: Line $l$ and $l_2$ are coplanar lines.
Statement $2$: Line $l$ and $l_2$ are intersecting lines.

Let the image of the point $P(1, 2, 3)$ in the plane $2x - y + z = 9$ be $Q$. If the coordinates of the point $R$ are $(6, 10, 7)$,then the square of the area of the triangle $PQR$ is $.....$.