Question

If $P$ be the point $(2,6,3),$ then the equation of the plane through $P$, at right angles to $OP$, where $O$ is the origin is:

• A. $2x+6y+3z=7$
• B. $2x-6y+3z=7$
• C. $2x+6y-3z=49$
• D. $2x+6y+3z=49$

Answer 1

The correct option is D $2x+6y+3z=49$

$D.c^{\prime }s$ of normal to the plane is $(\frac{2}{7},\frac{6}{7},\frac{3}{7})$
and perpendicular distance from origin to the plane $=\sqrt{4+36+9}=7$
So, using normal form of plane:
$lx+my+nz=p\Rightarrow \frac{2x}{7}+\frac{6y}{7}+\frac{3z}{7}=7\Rightarrow 2x+6y+3z=49$