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(B) Let the equation of the plane be $ax + by + cz = d$. Since it passes through $(0, 0, 0)$,we have $a(0) + b(0) + c(0) = d$,so $d = 0$. Thus,the pla

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$x - 19y - 11z = 0$

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(B) Let the equation of the plane be $ax + by + cz = d$. Since it passes through $(0, 0, 0)$,we have $a(0) + b(0) + c(0) = d$,so $d = 0$. Thus,the plane is $ax + by + cz = 0$. Since it passes through $(3, -1, 2)$,we have $3a - b + 2c = 0$ (Equation $1$). The plane is parallel to the line with direction ratios $(1, -4, 7)$,so the normal to the plane is perpendicular to the line. Thus,$a(1) + b(-4) + c(7) = 0$,which gives $a - 4b + 7c = 0$ (Equation $2$). Solving Equations $1$ and $2$ using cross-multiplication: $\frac{a}{(-1)(7) - (2)(-4)} = \frac{-b}{(3)(7) - (2)(1)} = \frac{c}{(3)(-4) - (-1)(1)}$ $\frac{a}{-7 + 8} = \frac{-b}{21 - 2} = \frac{c}{-12 + 1}$ $\frac{a}{1} = \frac{-b}{19} = \frac{c}{-11}$ So,$a = 1, b = -19, c = -11$. The equation of the plane is $1x - 19y - 11z = 0$.

The equation of the plane which is parallel to the line $\frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7}$ and passes through the points $(0, 0, 0)$ and $(3, -1, 2)$ is

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