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(C) The normal vector to the plane $x-y+z=3$ is $\vec{n} = (1, -1, 1)$.Let $Q = (3, 1, 7)$. The line passing through $Q$ and perpendicular to the plan

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$x-4y+7z=0$

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(C) The normal vector to the plane $x-y+z=3$ is $\vec{n} = (1, -1, 1)$.Let $Q = (3, 1, 7)$. The line passing through $Q$ and perpendicular to the plane is given by $\frac{x-3}{1} = \frac{y-1}{-1} = \frac{z-7}{1} = \lambda$.Any point on this line is $(\lambda+3, -\lambda+1, \lambda+7)$.For the intersection point $M$ with the plane,we have $(\lambda+3) - (-\lambda+1) + (\lambda+7) = 3$,which simplifies to $3\lambda + 9 = 3$,so $3\lambda = -6$,giving $\lambda = -2$.Thus,$M = (-2+3, -(-2)+1, -2+7) = (1, 3, 5)$.Since $M$ is the midpoint of $PQ$,if $P = (a, b, c)$,then $\frac{3+a}{2} = 1, \frac{1+b}{2} = 3, \frac{7+c}{2} = 5$.This gives $a = -1, b = 5, c = 3$,so $P = (-1, 5, 3)$.The plane passes through $P(-1, 5, 3)$ and contains the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$,which passes through the origin $O(0, 0, 0)$ with direction vector $\vec{v} = (1, 2, 1)$.The normal to the required plane is $\vec{n'} = \vec{OP} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 5 & 3 \ 1 & 2 & 1 \end{vmatrix} = \hat{i}(5-6) - \hat{j}(-1-3) + \hat{k}(-2-5) = -\hat{i} + 4\hat{j} - 7\hat{k}$.The equation of the plane is $-1(x-0) + 4(y-0) - 7(z-0) = 0$,which is $-x + 4y - 7z = 0$,or $x - 4y + 7z = 0$.

Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

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