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The perpendicular distance of the point $(1,-1,2)$ from the plane $x+2 y+z=4$, is
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Verified Answer
The correct answer is:
$\sqrt{\frac{3}{2}}$
Perpendicular distance of point $\left(x_1, y_1, z_1\right)$ from the plane $a x+b y+c z+d=0$ is given by $\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|$
$$
\therefore \text { Required distance }=\left|\frac{1-2+2-4}{\sqrt{(1)^2+(2)^2+(1)^2}}\right|=\frac{3}{\sqrt{6}}=\sqrt{\frac{3}{2}}
$$
$$
\therefore \text { Required distance }=\left|\frac{1-2+2-4}{\sqrt{(1)^2+(2)^2+(1)^2}}\right|=\frac{3}{\sqrt{6}}=\sqrt{\frac{3}{2}}
$$
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