Question

Let $\frac{ x -2}{3}=\frac{ y +1}{-2}=\frac{ z +3}{-1}$ lie on the plane $px - qy + z =5$, for some $p , q \in R$. The shortest distance of the plane from the origin is:

• A. $\sqrt{\frac{3}{109}}$
• B. $\sqrt{\frac{5}{142}}$
• C. $\sqrt{\frac{5}{71}}$
• D. $\sqrt{\frac{1}{142}}$

Answer 1

Answer: (B)

$(2, -1,-3)$ satisfy the given plane.
So $2 p + q =8$
Also given line is perpendicular to normal plane so
$3 p+2 q-1=0$
$\Rightarrow p =15, q =-22$
Eq. of plane $15 x-22 y+z-5=0$
its distance from origin $=\frac{6}{\sqrt{710}}=\sqrt{\frac{5}{142}}$