A plane passes through the point A ≡ (1, 2, 3). If the distance of this plane from the origin is maximum, then it’s equation is:
Answer & explanation
Correct answer: option 2
Clearly in this case OA will be a normal to the plane.
Direction cosine of segment OA are
$\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$
and $OA=\sqrt{1+4+0}=\sqrt{14}$
Thus the equation of plane is
x + 2y + 3z = 14