The equation of the plane passing through the point (−1, −1, 2) is:
a(x + 1) + b(y + 1) + c (z – 2) = 0 ...(1)
Where a, b and c are the direction ratios of the normal to the plane.
It is given that the plane (1) is perpendicular to the planes.
2x +3y – 3z = 2 and 5x – 4y + z = 6
∴2a + 3b – 3c = 0 ...(2)
5a – 4b + c = 0 ...(3)
Solving equations (2) and (3), we have:

a

3×1−(−4×−3)

=

b

−3×5−2×1

=

c

2(−4)−3×5

⇒

a

−9

=

b

−17

=

c

−23

So the direction ratios of the normal to the required plane are multiples of 9, 17, and 23.
Thus, the equation of the required plane is:
9 x + 1 + 17 y + 1 + 23 z - 2 = 0
or 9x + 17y + 23z = 20
OR
Equation of the plane passing through the point (3, 4, 1) is:
a (x - 3) + b (y - 4) + c (z - 1) = 0 ... 1
Where a, b, c are the direction ratios of the normal to the plane
It is given that the plane (1) passes through the point (0, 1, 0).
∴ a - 3 + b - 3 + c - 1 = 0
⇒ 3a + 3b + c = 0 ... 2
It is also given that the plane (1) is parallel to the line

x+3

2

=

y−3

7

=

z−2

5

So, this line is perpendicular to the normal of the plane (1).
∴ 2a + 7b + 5c = 0 ... 3
Solving equations (2) and (3), we have:

a

5×3−7×1

=

b

1×2−5×3

=

c

3×7−2×3

⇒

a

8

=

b

−13

=

c

15

So, the direction ratios of the normal to the required plane are multiples of 8, −13, 15.
Therefore, equation (1) becomes:
8 x - 3 - 13 y - 4 + 15 z - 1 = 0
⇒ 8x - 13y + 15z + 13 = 0, which is the required equation of the plane.