$A$ plane passes through $(2,3,-1)$ and is perpendicular to the line having direction ratios $3,-4,7$. The perpendicular distance from the origin to this 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:

$A$ plane $P$ is parallel to two lines whose direction ratios are $-2, 1, -3$ and $-1, 2, -2$,and it contains the point $(2, 2, -2)$. Let $P$ intersect the coordinate axes at the points $A, B, C$ making the intercepts $\alpha, \beta, \gamma$. If $V$ is the volume of the tetrahedron $OABC$,where $O$ is the origin and $p = \alpha + \beta + \gamma$,then the ordered pair $(V, p)$ is equal to.

The foot of the perpendicular drawn from the point $(-2,-1,3)$ to a plane $\pi$ is $(1,0,-2)$. If $a, b, c$ are the intercepts made by the plane $\pi$ on $X, Y, Z$-axes respectively, then $3a+b+5c=$

Let $6x - 3y + 2z - 6 = 0$ be the given plane. If $a, b, c$ are the intercepts made by the plane on $X, Y, Z$-axes respectively; $l, m, n$ are the direction cosines of a normal drawn to the plane and $p$ is the perpendicular distance from the origin to the plane, then $|al + bm + cn|=$

The Cartesian equation of the plane $\vec{r}=(2 \hat{i}-3 \hat{j})+\lambda(\hat{i}+2 \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$ is