Question

If $5,7,6$ are the sums of the $x,y$ intercepts; $y,z$ intercepts, $z,x$ intercepts respectively of a plane then the perpendicular distance from the origin to that plane is

• A. $\frac{144}{61}$
• B. $\frac{12}{\sqrt{61}}$
• C. $\frac{\sqrt{61}}{12}$
• D. $\frac{61}{144}$

Answer 1

The correct option is B $\frac{12}{\sqrt{61}}$

Let equation of plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
So intercepts on the axes are $a,b,c$ respectively.
Given $a+b=5,b+c=7,c+a=6$
$\Rightarrow a=2,b=3,c=4$
Therefore plane is $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1\Rightarrow 6x+4y+3z-12=0$

Hence, distance of this plane from origin is
$=\left|\frac{-12}{6^2+4^2+3^2}\right|=\frac{12}{\sqrt{61}}$