The equation of plane passing through a point $A$ $(2,-1,3)$ and parallel to the vectors $\mathbf{a}=(3,0,-1)$ and $\mathbf{b}=(-3,2,2)$ is

Since, plane is parallel to a given vector, it implies that normal of plane must perpendicular to the given vectors. Given point to which plane passes through is $(2,-1,3)$. Let $A, B$ and $C$ are direction ratios of its normal. $\therefore$ Equation of plane is $A(x-2)+B(y+1)+C(z-3)=0$ Now, normal to plane $A \mathbf{i}+B \mathbf{j}+C \mathbf{k}$ is perpendicular to the given vectors and $\quad \begin{aligned} & \mathbf{a}=3 \mathbf{i}+0 \mathbf{j}-\mathbf{k} \\ & \mathbf{b}=-3 \mathbf{i}+2 \mathbf{j}+2 \mathbf{k}\end{aligned}$ $\therefore \quad 3 A+0 B-C=0$..(i) and $\quad-3 A+2 B+2 C=0$..(ii) From Eqs. (i) and (ii), we get $\frac{A}{2}=\frac{B}{-3}=\frac{C}{6}$ $\therefore$ Equation of plane be $\begin{array}{ll} & 2(x-2)-3(y+1)+6(z-3)=0 \\ \text { i.e., } \quad & 2 x-3 y+6 z-25=0\end{array}$