The equation of the plane containing the line | vedclass

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(C) The equation of a plane passing through the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is given by $a(x + 1) + b(y - 3) + c(z + 2

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$x + y + z = 0$

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(C) The equation of a plane passing through the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is given by $a(x + 1) + b(y - 3) + c(z + 2) = 0$,where $-3a + 2b + c = 0$ (since the normal vector is perpendicular to the line direction vector $(-3, 2, 1)$).Since the plane passes through the point $(0, 7, -7)$,we substitute these coordinates into the plane equation:$a(0 + 1) + b(7 - 3) + c(-7 + 2) = 0$$a + 4b - 5c = 0$.Now we have the system of equations:$1) -3a + 2b + c = 0$$2) a + 4b - 5c = 0$Using cross-multiplication to find the ratios of $a, b, c$:$\frac{a}{(2)(-5) - (1)(4)} = \frac{b}{(1)(1) - (-3)(-5)} = \frac{c}{(-3)(4) - (2)(1)}$$\frac{a}{-10 - 4} = \frac{b}{1 - 15} = \frac{c}{-12 - 2}$$\frac{a}{-14} = \frac{b}{-14} = \frac{c}{-14}$This simplifies to $a = b = c$. Let $a = b = c = 1$.The equation of the plane is $1(x + 1) + 1(y - 3) + 1(z + 2) = 0$,which simplifies to $x + y + z = 0$.

The equation of the plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$ is

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