Let the equation of the plane be,
A (x−x1) + B (y−y1) + C (z−z1) = 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,

A

2×3−9×3

=

B

9×3−1×3

=

C

1×9−2×9

⇒

A

6−27

=

B

27−3

=

C

9−18

⇒

A

−21

=

B

24

=

C

−9

⇒

A

7

=

B

−8

=

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