Find the distance of the point (2, 3, -5) fro | vedclass

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(C) 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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$3$

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(C) 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 $(x_1, y_1, z_1) = (2, 3, -5)$ and the plane equation $x + 2y - 2z - 9 = 0$,we have $A = 1, B = 2, C = -2, D = -9$.Substituting these values into the formula:$d = \left| \frac{1(2) + 2(3) - 2(-5) - 9}{\sqrt{1^2 + 2^2 + (-2)^2}} \right|$$d = \left| \frac{2 + 6 + 10 - 9}{\sqrt{1 + 4 + 4}} \right|$$d = \left| \frac{9}{\sqrt{9}} \right|$$d = \frac{9}{3} = 3$Thus,the distance is $3$ units.

Find the distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$.

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