The Cartesian equation of the plane whose vector equation is $\vec{r}=(1+\lambda-\mu) \hat{i}+(2-\lambda) \hat{j}+(3-2 \lambda+2 \mu) \hat{k}$,where $\lambda, \mu$ are scalars,is:

The plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$. Prove that the equation of the plane in its new position is $ax + by \pm (\sqrt{a^{2} + b^{2}} \tan \alpha) z = 0$.

If $\bar{r}=(2-\lambda+\mu) \hat{i}+(1-\mu) \hat{j}+(2-3 \lambda+2 \mu) \hat{k}$ is the vector equation of a plane, then the equivalent cartesian equation of the plane is

If the foot of the perpendicular drawn from the origin to a plane is $(1, 2, 3)$,then a point on that plane is

Consider the following three planes:
$P : x + y - 2z + 7 = 0$
$Q : x + y + 2z + 2 = 0$
$R : 3x + 3y - 6z - 11 = 0$

Let $P$ be the plane passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ is