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Foot of perpendicular from point (3, 1, 2) to plane x + y + z = 6:
A(0, 0, 0)
BCannot determine
C(4, 2, 3) (using F = P + (d/|n|²) n where d = 6 - (3+1+2) = 0)
D(3, 1, 2) — point is on plane
Answer & Solution
Correct answer: D. (3, 1, 2) — point is on plane
Check: 3 + 1 + 2 = 6 = RHS. So the point (3, 1, 2) is ON the plane. Foot of perpendicular = point itself. (If not on plane, F = P - ((aP_x + bP_y + cP_z - d)/(a² + b² + c²)) × (a, b, c).)
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