Find the distance from the point (3, 2, 2) to the plane $-2x + 4y - 4z = 4$.
Added by Randy W.
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The plane is $-2x + 4y - 4z = 4$, which can be rewritten as $-2x + 4y - 4z - 4 = 0$. Thus, $A = -2$, $B = 4$, $C = -4$, and $D = -4$. Show more…
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