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(A) The equation of the plane passing through the point $(-1, 3, 2)$ is given by $a(x + 1) + b(y - 3) + c(z - 2) = 0$  $(1)$.Where $a, b, c$ are the d

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$7x - 8y + 3z + 25 = 0$

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(A) The equation of the plane passing through the point $(-1, 3, 2)$ is given by $a(x + 1) + b(y - 3) + c(z - 2) = 0$  $(1)$.Where $a, b, c$ are the direction ratios of the normal to the plane.Since the plane is perpendicular to $x + 2y + 3z = 5$,we have $a(1) + b(2) + c(3) = 0$,which implies $a + 2b + 3c = 0$  $(2)$.Since the plane is also perpendicular to $3x + 3y + z = 0$,we have $a(3) + b(3) + c(1) = 0$,which implies $3a + 3b + c = 0$  $(3)$.Solving equations $(2)$ and $(3)$ using cross-multiplication:$\frac{a}{(2)(1) - (3)(3)} = \frac{b}{(3)(3) - (1)(1)} = \frac{c}{(1)(3) - (2)(3)}$$\frac{a}{2 - 9} = \frac{b}{9 - 1} = \frac{c}{3 - 6}$$\frac{a}{-7} = \frac{b}{8} = \frac{c}{-3} = k$Thus,$a = -7k, b = 8k, c = -3k$.Substituting these into $(1)$:$-7k(x + 1) + 8k(y - 3) - 3k(z - 2) = 0$$-7x - 7 + 8y - 24 - 3z + 6 = 0$$-7x + 8y - 3z - 25 = 0$Multiplying by $-1$,we get $7x - 8y + 3z + 25 = 0$.

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 $3x + 3y + z = 0$.

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