Let the plane passing through the point $(2,1,-1)$ and containing the line joining the points $(1,3,2)$ and $(1,2,1)$ make intercepts $p, q, r$ on the coordinate axes. Then $p+q+r=$

The equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$ is:

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 $p - q =$

Let $\alpha, \beta, \gamma, \delta$ be real numbers such that $\alpha^2+\beta^2+\gamma^2 \neq 0$ and $\alpha+\gamma=1$. Suppose the point $(3,2,-1)$ is the mirror image of the point $(1,0,-1)$ with respect to the plane $\alpha x+\beta y+\gamma z=\delta$. Then which of the following statements is/are $TRUE$?
$(A)$ $\alpha+\beta=2$
$(B)$ $\delta-\gamma=3$
$(C)$ $\delta+\beta=4$
$(D)$ $\alpha+\beta+\gamma=\delta$

The equation of the plane,passing through the point $(-1, 2, -3)$ and parallel to the lines $\frac{x-1}{3} = \frac{y-2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y-1}{-3} = \frac{z-2}{2}$,is

If the plane $7x + 11y + 13z = 3003$ meets the coordinate axes at $A, B, C$,then the centroid of the $\triangle ABC$ is