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:

x + 2y + 3z = 0

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