The distance of a point (1, 2, -1) from the p | 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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$\sqrt{\frac{3}{7}}$ 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: $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$ Given the point $(1, 2, -1)$ and the plane $x - 2y + 4z + 10 = 0$,we have $A=1, B=-2, C=4, D=10$ and $x_1=1, y_1=2, z_1=-1$. Substituting these values into the formula: $d = \frac{|1(1) - 2(2) + 4(-1) + 10|}{\sqrt{1^2 + (-2)^2 + 4^2}}$ $d = \frac{|1 - 4 - 4 + 10|}{\sqrt{1 + 4 + 16}}$ $d = \frac{|3|}{\sqrt{21}} = \frac{3}{\sqrt{21}} = \frac{3}{\sqrt{3 	imes 7}} = \frac{\sqrt{3}}{\sqrt{7}} = \sqrt{\frac{3}{7}}$ units. Thus,the correct option is $D$.

The distance of a point $(1, 2, -1)$ from the plane $x - 2y + 4z + 10 = 0$ is

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