Find the Cartesian equation of the following plane:
$\vec{r} \cdot [(s-2t) \hat{i} + (3-t) \hat{j} + (2s+t) \hat{k}] = 15$

The plane which bisects the line segment joining the points $(-3, -3, 4)$ and $(3, 7, 6)$ at right angles,passes through which one of the following points?

$A$ variable plane passes through a fixed point $(\alpha, \beta, \gamma)$ and meets the coordinate axes in $A, B$ and $C$. Let $P_1, P_2$ and $P_3$ be the planes passing through $A, B, C$ and parallel to the coordinate planes $YZ, ZX, XY$ respectively. Then,the locus of the point of intersection of the planes $P_1, P_2$ and $P_3$ is

The distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is given by $\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$. If the plane $2x-y+2z+3=0$ is at distances of $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x-2y+4z+\lambda=0$ and $2x-y+2z+\mu=0$ respectively,then the maximum value of $\lambda+\mu$ is:

$A$ plane meets the coordinate axes at $P, Q, R$ respectively. If the centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$, then the equation of the plane is

Find the vector equation of a plane which is at a distance of $7$ units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$.