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If the foot of the perpendicular drawn from the origin to a plane is $(1,2,3)$, then a point on that plane is
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Verified Answer
The correct answer is:
$(7,2,1)$

DR′s of OP = < 1 − 0, 2 − 0, 3 − 0 > = < 1, 2, 3>
Since, OP is per pen dic u lar to the plane, there fore
OP is nor mal to the plane.
∴ Equation of plane pass ing through (1, 2, 3) and
hav ing d′r < 1, 2, 3> of its normal is
1(x − 1) + 2(y − 2) + 3(z − 3) = 0
⇒ x + 2y + 3z − 14 = 0
∴ (7, 2, 1) lies on the given plane.
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