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Find the distance $D$ between the point $(1,-4,-3)$ and the plane $2 x-3 y+6 z=-1$.
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Solution

Since the distance formulas require that the equations of the line and plane be written with zero on the right side, we first need to rewrite the equation of the plane as
$$2 x-3 y+6 z+1=0$$
from which we obtain
$$D=\frac{|2(1)+(-3)(-4)+6(-3)+1|}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}=\frac{|-3|}{7}=\frac{3}{7}$$

by Platinum (132,156 points)

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