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(C) The perpendicular distance of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by the formula:$D = \left| \frac{ax_1 + by_

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$\sqrt{\frac{3}{2}}$

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(C) The perpendicular distance of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by the formula:$D = \left| \frac{ax_1 + by_1 + cz_1 + d}{\sqrt{a^2 + b^2 + c^2}} ight|$Given the point $(1, -1, 2)$ and the plane $x + 2y + z - 4 = 0$, we have $a=1, b=2, c=1, d=-4$.Substituting these values into the formula:$D = \left| \frac{1(1) + 2(-1) + 1(2) - 4}{\sqrt{1^2 + 2^2 + 1^2}} ight|$$D = \left| \frac{1 - 2 + 2 - 4}{\sqrt{1 + 4 + 1}} ight|$$D = \left| \frac{-3}{\sqrt{6}} ight| = \frac{3}{\sqrt{6}}$Rationalizing the expression:$D = \frac{3}{\sqrt{6}} 	imes \frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}$

The perpendicular distance of the point $(1, -1, 2)$ from the plane $x + 2y + z = 4$ is

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