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: $\frac{a}{3\times 1 - (-4\times -3)}$ = $\frac{b}{-3\times 5 - 2\times 1}$ = $\frac{c}{2(-4) - 3\times 5}$ ⇒ $\frac{a}{-9}$ = $\frac{b}{-17}$ = $\frac{c}{-23}$ 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 $\frac{x+3}{2}$ = $\frac{y-3}{7}$ = $\frac{z-2}{5}$ So, this line is perpendicular to the normal of the plane (1). ∴ 2a + 7b + 5c = 0 ... 3 Solving equations (2) and (3), we have: $\frac{a}{5\times 3 - 7\times 1}$ = $\frac{b}{1\times 2 - 5\times 3}$ = $\frac{c}{3\times 7 - 2\times 3}$ ⇒ $\frac{a}{8}$ = $\frac{b}{-13}$ = $\frac{c}{15}$ 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.