The Cartesian equation of the plane passing through the point $A(7,8,6)$ and parallel to the $XY$ plane is

The angle between the planes $x+y+2z=6$ and $2x-y+z=9$ is:

Let $(\alpha, \beta, \gamma)$ be the image of the point $P (2, 3, 5)$ in the plane $2x + y - 3z = 6$. Then $\alpha + \beta + \gamma$ is equal to

$A$ plane passes through $(2,1,2)$ and $(1,2,1)$ and is parallel to the line $2x = 3y$ and $z = 1$. Then the plane also passes through which of the following points?

If the vector equation of the plane $\bar{r}=(2 \hat{i}+\hat{k})+\lambda \hat{i}+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ in scalar product form is given by $\bar{r} \cdot(3 \hat{j}+2 \hat{k})=\alpha$,then $\alpha=$

The perpendicular distance of the origin from the plane $x-3y+4z-6=0$ is