The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $X$-axis at $A$,$Y$-axis at $B$,and $Z$-axis at $C$. The area of $\triangle ABC$ is:

Consider three planes:
$P_1: x-y+z=1$
$P_2: x+y-z=-1$
$P_3: x-3y+3z=2$
Let $L_1, L_2, L_3$ be the lines of intersection of the planes $P_2$ and $P_3$,$P_3$ and $P_1$,and $P_1$ and $P_2$,respectively.
$STATEMENT-1$: At least two of the lines $L_1, L_2$ and $L_3$ are non-parallel.
$STATEMENT-2$: The three planes do not have a common point.

$\vec{n}$ is a unit vector normal to the plane $\pi$ containing the vectors $\hat{i}+3 \hat{k}$ and $2 \hat{i}+\hat{j}-\hat{k}$. If this plane $\pi$ passes through the point $(-3,7,1)$ and $p$ is the perpendicular distance from the origin to this plane $\pi$,then $\sqrt{p^2+5}=$

The direction ratios of the normal to the plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle $\frac{\pi}{4}$ with the plane $x + y = 3$ are:

$A$ vector $\vec{n}$ is inclined to $X$-axis at $45^{\circ}$,$Y$-axis at $60^{\circ}$ and at an acute angle to $Z$-axis. If $\vec{n}$ is normal to a plane passing through the point $(-\sqrt{2}, 1, 1)$,then the equation of the plane is

The equation of a plane containing the point $(1, -1, 1)$ and parallel to the plane $2x + 3y - 4z = 17$ is