Suppose the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta=0$. Then $(\alpha+\beta)$ is equal to ... .

The line passing through $(4, -1, 2)$ and $(-3, 2, 3)$ meets the plane at right angles at the point $(-10, 5, 4)$. Then the equation of the plane is:

If the line joining the points $A(1,0,0)$ and $B(0,0,1)$ is a normal to the plane $\pi$ which passes through the point $A$, then the angle between the planes $\pi$ and $x+y+z=6$ is

If $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+4 \bar{k})=5$ and $\bar{r} \cdot(\bar{i}+\bar{j}-\bar{k})=7$ are two planes and $(16, -9, 0)$ is a point common to both the planes, then the vector equation of the line of intersection of the planes is $\bar{r}=$

Find the equation of the plane which passes through the points $(0,1,2)$ and $(-1,0,3)$ and is perpendicular to the plane $2x+3y+z=5$.

The line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}$ meets the plane $x + 2y + 3z = 14$ at the point: