The equation of the plane which is bisecting the line segment joining the points $A(2,3,4)$ and $B(-4,1,-2)$ and is perpendicular to it,is

The distance of the point $(5,3,-1)$ from the plane passing through the points $A(2,1,0)$,$B(3,-2,4)$,and $C(1,-3,3)$ is:

$A$ plane $\pi$ passing through the point $3 \hat{i}-4 \hat{j}+5 \hat{k}$ is parallel to the plane which passes through the point $\hat{i}+\hat{j}-\hat{k}$ and is perpendicular to the vector $\hat{i}+2 \hat{j}-3 \hat{k}$. Then the Cartesian equation of $\pi$ is

The equation of the plane passing through $(1, 1, 1)$ and $(1, -1, -1)$ and perpendicular to $2x - y + z + 5 = 0$ is:

$P$ and $Q$ are points on the straight line passing through the point $A(3 \hat{i}+\hat{j}-\hat{k})$ and parallel to the vector $\vec{v} = 2 \hat{i}-\hat{j}+2 \hat{k}$. If $AP = AQ = 3$,then the vector equation of the plane $OPQ$ is:

The direction cosines of the normal to the plane containing the lines having direction ratios $1, 2, 1$ and $4, 5, -3$ are