Question

The equation of plane passing through a point $A(2,-1,3)$ and parallel to the vectors $a=(3,0,-1)andb=(-3,2,2)$ is:

Answer 1

Let the direction ratios of the normal of the required plane be $a,b,c.$

The plane is parallel to $(3,0,-1),$ hence, the normal must be perpendicular.

$3a+0b–c=0\dots \left(1\right)$

It is also parallel to$(-3,2,2),so,–3a+2b+2c=0\dots \left(2\right)$

From $\left(1\right)and\left(2\right)$

$a=2,b=–3,c=6$

It passes through$(2,–1,3).$

$$\begin{gathered} 2(x–2)–3(y+1)+6(z–3)=0 \\ \Rightarrow 2x–4–3y–3+6z–18=0 \\ \Rightarrow 2x–3y+6z–25=0 \end{gathered}$$

Hence, the required equation of the plane is $2x-3y+6z-25=0$