The distance of the point (2, 3, -5) from the | vedclass

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(B) 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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$3$

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(B) 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: $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$ Given the point $(2, 3, -5)$ and the plane $x + 2y - 2z - 9 = 0$,we have $A=1, B=2, C=-2, D=-9$. Substituting these values into the formula: $d = \frac{|(1)(2) + (2)(3) + (-2)(-5) - 9|}{\sqrt{1^2 + 2^2 + (-2)^2}}$ $d = \frac{|2 + 6 + 10 - 9|}{\sqrt{1 + 4 + 4}}$ $d = \frac{|9|}{\sqrt{9}}$ $d = \frac{9}{3} = 3$ Thus,the distance is $3$ units.

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

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