The equation of the plane passing through (-2 | vedclass

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Question

(B) The equation of a plane passing through $(-2, 2, 2)$ is given by $a(x + 2) + b(y - 2) + c(z - 2) = 0$. Since the plane passes through $(2, -2, -2)

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$5x + 3y + 2z = 0$

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(B) The equation of a plane passing through $(-2, 2, 2)$ is given by $a(x + 2) + b(y - 2) + c(z - 2) = 0$. Since the plane passes through $(2, -2, -2)$,we substitute these coordinates into the equation: $a(2 + 2) + b(-2 - 2) + c(-2 - 2) = 0 \Rightarrow 4a - 4b - 4c = 0 \Rightarrow a - b - c = 0$ (Equation $1$). The plane is perpendicular to $9x - 13y - 3z = 0$,so the normal vector $(a, b, c)$ is perpendicular to the normal vector $(9, -13, -3)$ of the given plane. Thus,$9a - 13b - 3c = 0$ (Equation $2$). Solving equations $(1)$ and $(2)$ using cross-multiplication: $\frac{a}{(-1)(-3) - (-1)(-13)} = \frac{-b}{(1)(-3) - (-1)(9)} = \frac{c}{(1)(-13) - (-1)(9)}$ $\frac{a}{3 - 13} = \frac{-b}{-3 + 9} = \frac{c}{-13 + 9}$ $\frac{a}{-10} = \frac{-b}{6} = \frac{c}{-4} \Rightarrow \frac{a}{5} = \frac{b}{3} = \frac{c}{2}$. Substituting the direction ratios $(5, 3, 2)$ into the plane equation: $5(x + 2) + 3(y - 2) + 2(z - 2) = 0$ $5x + 10 + 3y - 6 + 2z - 4 = 0$ $5x + 3y + 2z = 0$.

The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

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