The equation of the plane containing the line | vedclass

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(B) Let the equation of the plane be $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. Since the plane passes through the point $(-1, 3, -2)$ (from the line equat

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$x+y+z=0$

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(B) Let the equation of the plane be $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. Since the plane passes through the point $(-1, 3, -2)$ (from the line equation),we have $a(x+1) + b(y-3) + c(z+2) = 0$. Since the plane also passes through $(0, 7, -7)$,we substitute these coordinates: $a(0+1) + b(7-3) + c(-7+2) = 0$,which simplifies to $a + 4b - 5c = 0$. Also,the line lies in the plane,so the normal vector $(a, b, c)$ is perpendicular to the direction vector of the line $(-3, 2, 1)$. Thus,$-3a + 2b + c = 0$. Solving the system of equations $a + 4b - 5c = 0$ and $-3a + 2b + c = 0$ using cross multiplication: $\frac{a}{(4)(1) - (-5)(2)} = \frac{-b}{(1)(1) - (-5)(-3)} = \frac{c}{(1)(2) - (4)(-3)}$ $\frac{a}{4+10} = \frac{-b}{1-15} = \frac{c}{2+12} \Rightarrow \frac{a}{14} = \frac{b}{14} = \frac{c}{14}$. Taking $a=1, b=1, c=1$,the equation becomes $1(x+1) + 1(y-3) + 1(z+2) = 0$,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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