The ratio in which the line joining $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is

Let $P$ be the plane,which contains the line of intersection of the planes $x + y + z - 6 = 0$ and $2x + 3y + z + 5 = 0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to

Let $L$ be the line passing through the points $\hat{i}-9 \hat{k}$ and $7 \hat{j}+\hat{k}$ and $\pi$ be the plane passing through the point $6 \hat{i}+\hat{j}$ and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$. If $\theta$ is the angle between $L$ and $\pi$,then $\sin \theta=$

The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane

$A$ 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.

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^o$ about its line of intersection with the plane $5x - 4y - 2z + 1 = 0$. The equation of the plane in its new position is: