Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Solution:

It is known to us that the eq. of a plane passing through \left(x_{1}, y_{1}, z_{1}\right) is given by
A\left(x-x_{1}\right)+B\left(y-y_{1}\right)+C\left(z-z_{1}\right)=0
Where, A, B, C are the direction ratios of normal to the plane.
It is given that the plane passes through (-1,3,2)
Therefore, eq. of plane is given by
A(x+1)+B(y-3)+C(z-2)=0 \ldots \ldots \ldots(1)
As this plane is perpendicular to the given two planes. Therefore, their normal to the plane would be perpendicular to normal of both planes.
It is known to us that
\vec{a} \times \vec{b} is perpendicular to both \vec{a} and \vec{b}
Therefore, required normal is cross product of normal of planes
x+2 y+3 z=5 and 3 x+3 y+z=0
\begin{aligned} \text { Required Normal } &=\left|\begin{array}{ccc} \hat{1} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 3 & 1 \end{array}\right| \\ =\hat{i}[2(1)-3(3)]-\hat{j}[1(1)-3(3)]+\hat{k}[1(3)-3(2)] \\ =\hat{i}[2-9]-\hat{j}[1-9]+\hat{k}[3-6] \\ =-7 \hat{i}+8 \hat{j}-3 \hat{k} \end{aligned}
Therefore, the direction ratios are =-7,8,-3
\therefore A=-7, B=8, C=-3
Substituting the obtained values in eq.(1), we obtain
\begin{array}{l} A(x+1)+B(y-3)+C(z-2)=0 \\ -7(x+1)+8(y-3)+(-3)(z-2)=0 \\ -7 x-7+8 y-24-3 z+6=0 \end{array}
\begin{array}{l} -7 x+8 y-3 z-25=0 \\ 7 x-8 y+3 z+25=0 \end{array}
As a result, the eq. of the required plane is 7 x-8 y+3 z+25=0.