$A$ plane passes through the points $A (1, 2, 3)$,$B (2, 3, 1)$,and $C (2, 4, 2)$. If $O$ is the origin and $P$ is $(2, -1, 1)$,then the projection of $\overline{OP}$ on this plane is of length .... .

The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = q$ is given by $\frac{|\vec{a} \cdot \vec{m} - q|}{|\vec{m}|}$. If the distance of the point $\hat{i} + 2\hat{j} + 3\hat{k}$ from the plane $\vec{r} \cdot (2\hat{i} + 6\hat{j} - 9\hat{k}) = -1$ is $p$ and the distance of the origin from this plane is $q$,then $p - q =$

The equation of the plane passing through $(1, -1, 2)$ and perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$ is:

$A$ plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(2, 3, 5)$. Then,the equation of that plane is

If a plane cuts off intercepts $-6, 3, 4$ from the coordinate axes,then the length of the perpendicular from the origin to the plane is

The direction cosines of the normal to the plane $x + 2y - 3z + 4 = 0$ are