Question

The perpendicular distance of the origin from the plane which makes intercepts 12, 3 and 4 on x, y, z axes
respectively, is

• A. 13
• B. 11
• C. 17
• D. $\frac{6\sqrt{2}}{\sqrt{13}}$

Answer 1

The correct option is D $\frac{6\sqrt{2}}{\sqrt{13}}$

Let equation of plane be lx + my + nz = p
$or\frac{x}{\left(\frac{p}{l}\right)}+\frac{y}{\left(\frac{p}{m}\right)}+\frac{z}{\left(\frac{p}{n}\right)}=1$
According to question,
$\frac{p}{l}=12,\frac{p}{m}=3,\frac{p}{n}=4or\frac{p}{12}=l,\frac{p}{3}=m,\frac{p}{4}=nor\frac{p^2}{144}+\frac{p^2}{9}+\frac{p^2}{16}=l^2+m^2+n^2=1\Rightarrow p^2(\frac{1}{144}+\frac{1}{9}+\frac{1}{16})=1p^2\left(\frac{1+16+9}{144}\right)=1or{p}^2=\frac{144}{26}=\frac{72}{13}\therefore p=\frac{6\sqrt{2}}{\sqrt{13}}$