The equation of the plane passing through the point $(1,1,1)$ and perpendicular to the planes $2x-y-2z=5$ and $3x-6y+2z=7$ is

The foot of the perpendicular drawn from the point $(-2,-1,3)$ to a plane $\pi$ is $(1,0,-2)$. If $a, b, c$ are the intercepts made by the plane $\pi$ on $X, Y, Z$-axes respectively, then $3a+b+5c=$

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

If the foot of the perpendicular drawn from the origin to a plane is $P(2,-1,4)$,then the equation of the plane is

Find the vector and Cartesian equations of the plane that passes through the point $(1, 4, 6)$ and the normal vector to the plane is $\hat{i} - 2\hat{j} + \hat{k}$.

If the planes $\bar{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3$ and $\bar{r} \cdot(4 \hat{i}-\hat{j}+\mu \hat{k})=5$ are parallel,then $\lambda+\mu=$