If the points $A(2-x, 2, 2)$,$B(2, 2-y, 2)$,$C(2, 2, 2-z)$,and $D(1, 1, 1)$ are coplanar,then the locus of point $P(x, y, z)$ is

$A$ plane $\pi_1$ passing through the point $3 \hat{i}-7 \hat{j}+5 \hat{k}$ is perpendicular to the vector $\hat{i}+2 \hat{j}-2 \hat{k}$ and another plane $\pi_2$ passing through the point $2 \hat{i}+7 \hat{j}-8 \hat{k}$ is perpendicular to the vector $3 \hat{i}+2 \hat{j}+6 \hat{k}$. If $p_1$ and $p_2$ are the perpendicular distances from the origin to the planes $\pi_1$ and $\pi_2$ respectively, then $p_1-p_2=$

$A$ plane meets the $X, Y, Z$-axes in $A, B, C$ respectively. If the centroid of the $\triangle ABC$ is $(2, -3, 5)$,then the perpendicular distance from the origin to the given plane is:

$P_1$ and $P_2$ are two distinct and intersecting planes. Three non-collinear points lie on $P_1$ and another three non-collinear points lie on $P_2$ (none being on the line of intersection of the planes). Then the maximum number of tetrahedrons formed using these six points is:

The point which lies on the plane passing through the points $A(\hat{i}-2\hat{j}-3\hat{k})$, $B(3\hat{i}-\hat{j}+4\hat{k})$, and $C(-3\hat{i}+2\hat{j}-5\hat{k})$ is:

In a tetrahedron $LMNO$,edges $ML, MN$ and $MO$ are mutually perpendicular. If the lengths of the altitudes drawn from $O, L$ and $N$ to their opposite faces are $1, 2$ and $3$ units respectively,then the length of the altitude drawn from $M$ to the face $LNO$ is: