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(C) The line passes through the point $P(1, 2, 3)$ and has direction ratios $(5, 4, 5)$.Since the plane passes through the origin $(0, 0, 0)$ and the

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$x - 5y + 3z = 0$

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(C) The line passes through the point $P(1, 2, 3)$ and has direction ratios $(5, 4, 5)$.Since the plane passes through the origin $(0, 0, 0)$ and the point $P(1, 2, 3)$,the vector $\vec{OP} = \hat{i} + 2\hat{j} + 3\hat{k}$ lies in the plane.The direction vector of the line is $\vec{v} = 5\hat{i} + 4\hat{j} + 5\hat{k}$.The normal to the plane $\vec{n}$ is given by the cross product $\vec{OP} 	imes \vec{v}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 5 & 4 & 5 \end{vmatrix} = \hat{i}(10 - 12) - \hat{j}(5 - 15) + \hat{k}(4 - 10) = -2\hat{i} + 10\hat{j} - 6\hat{k}$.Dividing by $-2$,we get the normal vector $\vec{n}' = \hat{i} - 5\hat{j} + 3\hat{k}$.The equation of the plane passing through $(0, 0, 0)$ is $1(x - 0) - 5(y - 0) + 3(z - 0) = 0$,which simplifies to $x - 5y + 3z = 0$.

The equation of the plane passing through the origin and containing the line $\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{5}$ is:

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