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(C) The points $P(1, 2, -1)$ and $Q(2, -1, 3)$ lie on the same side of the plane $-x + y + z - 1 = 0$.The perpendicular distance of point $P$ from the

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$26$

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(C) The points $P(1, 2, -1)$ and $Q(2, -1, 3)$ lie on the same side of the plane $-x + y + z - 1 = 0$.The perpendicular distance of point $P$ from the plane is $\left|\frac{-(1) + (2) + (-1) - 1}{\sqrt{(-1)^{2} + 1^{2} + 1^{2}}}\right| = \left|\frac{-1}{\sqrt{3}}\right| = \frac{1}{\sqrt{3}}$.The perpendicular distance of point $Q$ from the plane is $\left|\frac{-(2) + (-1) + (3) - 1}{\sqrt{(-1)^{2} + 1^{2} + 1^{2}}}\right| = \left|\frac{-1}{\sqrt{3}}\right| = \frac{1}{\sqrt{3}}$.Since the perpendicular distances are equal,the line segment $PQ$ is parallel to the given plane. Therefore,the distance $d$ between the feet of the perpendiculars $M$ and $N$ is equal to the distance between points $P$ and $Q$.$d = |PQ| = \sqrt{(2 - 1)^{2} + (-1 - 2)^{2} + (3 - (-1))^{2}}$$d = \sqrt{1^{2} + (-3)^{2} + 4^{2}} = \sqrt{1 + 9 + 16} = \sqrt{26}$.Thus,$d^{2} = 26$.

Let $d$ be the distance between the foot of perpendiculars of the points $P(1, 2, -1)$ and $Q(2, -1, 3)$ on the plane $-x + y + z = 1$. Then $d^{2}$ is equal to

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