If the perpendicular distance from $(1, 2, 4)$ to the plane $2x + 2y - z + k = 0$ is $3$,then $k =$

Let the plane $\pi$ pass through the point $(1,0,1)$ and be perpendicular to the planes $2x+3y-z=2$ and $x-y+2z=1$. Let the equation of the plane passing through the point $(11,7,5)$ and parallel to the plane $\pi$ be $ax+by-z-d=0$. Then,$\frac{a}{b}+\frac{b}{d}=$

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

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

The plane $ax + by + cz = 1$ meets the coordinate axes at points $A, B$,and $C$. Find the centroid of the triangle $ABC$.

Find the angle between the two planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$ using the vector method.