Let the equation of the plane be,
A $\left(\mathit{x}-{\mathit{x}}_1\right)$ + B $\left(\mathit{y}-{\mathit{y}}_1\right)$ + C $\left(\mathit{z}-{\mathit{z}}_1\right)$ = 0
Plane passes through the point (-1, 3, 2)
∴ A (x + 1) + B (y - 3) + C (z - 2) = 0 ... (i)
Now applying the condition of perpendicularity to the plane (i) with planes
x + 2y + 3z = 5 and 3x + 3y + z = 0, we have
A + 2B + 3C = 0
3A + 3B + C = 0
Solving we get
A + 2B + 3C = 0
9A + 9B + 3C = 0
By cross multiplication, we have,
$\frac{\mathit{A}}{2\times 3-9\times 3}$ = $\frac{\mathit{B}}{9\times 3-1\times 3}$ = $\frac{\mathit{C}}{1\times 9-2\times 9}$
⇒ $\frac{\mathit{A}}{6-27}$ = $\frac{\mathit{B}}{27-3}$ = $\frac{\mathit{C}}{9-18}$
⇒ $\frac{\mathit{A}}{-21}$ = $\frac{\mathit{B}}{24}$ = $\frac{\mathit{C}}{-9}$
⇒ $\frac{\mathit{A}}{7}$ = $\frac{\mathit{B}}{-8}$ = $\frac{\mathit{C}}{3}$
⇒ A = 7λ ; B = - 8λ ; C = 3λ
By substituting A and C in equation (i), we get,
Substituting the values of A, B and C in equation (i), we have,
7λ (x + 1) - 8λ (y - 3) + 3λ (z - 2) = 0
⇒ 7x + 7 - 8y + 24 + 3z - 6 = 0
⇒ 7x - 8y + 3z + 25 = 0