Find the angle between the planes $x+2y+2z-5=0$ and $3x+3y+2z-8=0$.

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:

$A$ vector $\vec{n}$ is inclined to the $x$-axis at $45^\circ$,to the $y$-axis at $60^\circ$,and at an acute angle to the $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2}, -1, 1)$,then the equation of the plane is:

Let $Q$ be the foot of the perpendicular from the origin to the plane $4x - 3y + z + 13 = 0$ and $R$ be a point $(-1, 1, -6)$ on the plane. Then the length $QR$ is

$A$ point moves such that its distances from the points $(3, 4, -2)$ and $(2, 3, -3)$ are equal. What is the locus of the point?

Find the distance of the point $(3, -2, 1)$ from the plane $2x - y + 2z + 3 = 0$. (in $/3$)