$P$ is a fixed point $(a, a, a)$ on a line through the origin equally inclined to the axes. Then,any plane through $P$ perpendicular to $OP$ makes intercepts on the axes,the sum of whose reciprocals 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 $p - q =$

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, 5)$. Then,the Cartesian equation of the plane is:

Consider the three planes $P_{1}: 3x + 15y + 21z = 9$; $P_{2}: x - 3y - z = 5$; and $P_{3}: 2x + 10y + 14z = 5$. Then,which one of the following is true?

Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:

If $A$ and $B$ are the feet of the perpendiculars drawn from point $Q(a, b, c)$ to the planes $yz$ and $zx$ respectively,then the equation of the plane passing through the points $A, B$ and the origin $O(0, 0, 0)$ is $.........$