What is the distance of the point (2, 3, 4) f | vedclass

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(A) 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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$1$

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(A) 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 point $(x_1, y_1, z_1) = (2, 3, 4)$ and plane $3x - 6y + 2z + 11 = 0$.Here,$A = 3, B = -6, C = 2, D = 11$.Substituting the values into the formula:$d = \frac{|3(2) - 6(3) + 2(4) + 11|}{\sqrt{3^2 + (-6)^2 + 2^2}}$$d = \frac{|6 - 18 + 8 + 11|}{\sqrt{9 + 36 + 4}}$$d = \frac{|7|}{\sqrt{49}}$$d = \frac{7}{7} = 1$Therefore,the distance is $1$ unit.

What is the distance of the point $(2, 3, 4)$ from the plane $3x - 6y + 2z + 11 = 0$?

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