If $(3,4,-7)$ is the foot of the perpendicular drawn from the point $(-2,3,6)$ to the plane $\pi$,then the sum of the intercepts made by the plane $\pi$ on the $X$ and $Y$-axes is

$A$ plane meets the coordinate axes at $A, B, C$ and $(\alpha, \beta, \gamma)$ is the centroid of the triangle $ABC$. Then the equation of the plane is

Let $\pi_1$ be the plane passing through the point $2\hat{i}-\hat{j}+\hat{k}$ and perpendicular to the vector $a\hat{i}+2\hat{j}-3\hat{k}$, and $\pi_2$ be the plane passing through the point $\hat{i}+2\hat{j}-\hat{k}$ and perpendicular to the vector $\hat{i}-2\hat{j}+\hat{k}$. If $\theta$ is the angle between the planes $\pi_1$ and $\pi_2$ and $\cos \theta = -\sqrt{\frac{3}{7}}$, then the integral value of $a$ is:

The sum of intercepts of the plane $4x + 3y + 2z = 2$ on the coordinate axes is

The vector equation of the plane passing through the point $2\hat{i} - \hat{j} - 4\hat{k}$ and parallel to the plane $\vec{r} \cdot (4\hat{i} - 12\hat{j} - 3\hat{k}) - 7 = 0$ is:

The distance of the point $(2, -1, 0)$ from the plane $2x + y + 2z + 8 = 0$ is