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(B) Let $M$ be the midpoint of the line segment $PQ$.The coordinates of $M$ are given by $\left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{5+3}{2}\right) = (

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$x+y-z-1=0$

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(B) Let $M$ be the midpoint of the line segment $PQ$.The coordinates of $M$ are given by $\left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{5+3}{2}\right) = (2, 3, 4)$.The direction ratios of the line $PQ$ are $(3-1, 4-2, 3-5) = (2, 2, -2)$.Since the plane is perpendicular to $PQ$,the normal vector to the plane is $\vec{n} = 2\hat{i} + 2\hat{j} - 2\hat{k}$.The equation of the plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the values,we get $2(x-2) + 2(y-3) - 2(z-4) = 0$.Dividing by $2$,we get $(x-2) + (y-3) - (z-4) = 0$.$x + y - z - 2 - 3 + 4 = 0$.$x + y - z - 1 = 0$.

The equation of the plane,passing through the midpoint of the line segment joining the points $P(1, 2, 5)$ and $Q(3, 4, 3)$ and perpendicular to it,is

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