The plane $2x + 3y + 4z = 1$ meets the $X$-axis at $A$,the $Y$-axis at $B$,and the $Z$-axis at $C$. Then the centroid of $\triangle ABC$ is:

Find the equation of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z - 1 = 0$.

The angle between a normal to the plane $2x - y + 2z - 1 = 0$ and the $X$-axis is

$A$ plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

Let $(\alpha, \beta, \gamma)$ be the image of the point $P (2, 3, 5)$ in the plane $2x + y - 3z = 6$. Then $\alpha + \beta + \gamma$ is equal to

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 $p - q =$