The plane containing the point $(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is

Given that point $(3,2,0)$ lies on plane and line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ also lies on the plane. $\because$ Plane contains points $(3,2,0)$ and $(3,6,4)$ The DR's of line joining these two points is $(0,4,4)$. Therefore, DR's of normal to plane is $(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \times(4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ Therefore, the equation of plane is $4 x-4 y+4 z=k$ $\because$ Point $(3,2,0)$ lies on planes. So, we get $\therefore \quad k=4 \times 3-4 \times 2+4 \times 0=4$ Hence, the equation of plane is $x-y+z=1$.