The distance of the point (2, -1, 0) from the | vedclass

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(D) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula:$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\

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$\frac{11}{3}$ units

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(D) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula:$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$Here,the point is $(2, -1, 0)$ and the plane is $2x + y + 2z + 8 = 0$.Substituting the values into the formula:$d = \frac{|2(2) + 1(-1) + 2(0) + 8|}{\sqrt{2^2 + 1^2 + 2^2}}$$d = \frac{|4 - 1 + 0 + 8|}{\sqrt{4 + 1 + 4}}$$d = \frac{|11|}{\sqrt{9}}$$d = \frac{11}{3}$ units.

The distance of the point $(2, -1, 0)$ from the plane $2x + y + 2z + 8 = 0$ is

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