The equation of the plane passing through the points $(0, 1, 0)$ and $(3, 4, 1)$ and parallel to the line $\frac{x + 3}{2} = \frac{y - 3}{7} = \frac{z - 2}{5}$ is

4 x - 13 y + 5 z - 13 = 0

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8 x - 13 y + 5 z + 15 = 0

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8 x - 13 y + 15 z + 13 = 0

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4 x - 3 y + 5 z + 15 = 0

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Solution

The correct option is C $8 x - 13 y + 15 z + 13 = 0$
$\begin{matrix} a (x) + b (y - 1) + c z = 0 3 a + 3 b + c = 0 2 a + 7 b + 5 c = 0 \frac{a}{8} = \frac{- b}{13} = \frac{c}{15} 8 x - 13 (y - 1) + 15 z = 0 8 x - 13 y + 15 z + 13 = 0 H e n c e, t h e o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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