Question

Let position vectors of points $A,B$ and $C$ of triangle $ABC$ respectively be $\hat{i}+\hat{j}+2\hat{k},$ $\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}+\hat{k}.$ Let $l_1,l_2,l_3$ be the lengths of perpendiculars drawn from the orthocentre $O$ on the sides $AB,$ $BC$ and $CA$ respectively. Then the value of $4(l_1+l_2+l_3{)}^2$ is

Answer 1

$AB=BC=CA=\sqrt{2}$
$ABC$ is an equilateral triangle.
$\therefore l_1=l_2=l_3=$ inradius $\left(r\right)$
$r=\frac{\Delta }{s}=\frac{\frac{\sqrt{3}}{4}(\sqrt{2}{)}^2}{\frac{3}{2}\left(\sqrt{2}\right)}=\frac{1}{\sqrt{6}}$
$\Rightarrow l_1+l_2+l_3=\frac{3}{\sqrt{6}}=\frac{\sqrt{6}}{2}$
$\therefore 4(l_1+l_2+l_3{)}^2=4\times \frac{6}{4}=6$