Find the distance of the point (3, -2, 1) fro | vedclass

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

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$13$

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(A) The distance $d$ of a point $P(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula:$d = \left| \frac{Ax_1 + By_1 + Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}} \right|$Given the point $(x_1, y_1, z_1) = (3, -2, 1)$ and the plane equation $2x - y + 2z + 3 = 0$,we have $A = 2, B = -1, C = 2, D = 3$.Substituting these values into the formula:$d = \left| \frac{2(3) + (-1)(-2) + 2(1) + 3}{\sqrt{2^2 + (-1)^2 + 2^2}} \right|$$d = \left| \frac{6 + 2 + 2 + 3}{\sqrt{4 + 1 + 4}} \right|$$d = \left| \frac{13}{\sqrt{9}} \right|$$d = \frac{13}{3}$

Find the distance of the point $(3, -2, 1)$ from the plane $2x - y + 2z + 3 = 0$. (in $/3$)

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