Question

Let $A,B,C$ be three points on $\overline{OX},\overline{OY},\overline{OZ}$ respectively at the distances $3,6,9$ from origin. Let $Q$ be the point $(2,5,8)$ and $P$ be the point equidistant from $O,A,B,C.$ Then, the coordinates of the point $R$ which divides $PQ$ in the ration $3:2$ is

• A. $\left(\frac{17}{10},\frac{29}{5},\frac{43}{10}\right)$
• B. $\left(\frac{7}{5},\frac{16}{5},5\right)$
• C. $\left(\frac{9}{5},\frac{21}{5},\frac{33}{5}\right)$
• D. $\left(\frac{8}{5},\frac{19}{5},6\right)$

Answer 1

Answer: (C)

Let $P$ is $(u,v,w)$
Here, $P{O}^2=P{A}^2,P{O}^2=P{B}^2,P{O}^2=P{C}^2$
$u^2+v^2+w^2=(u-3{)}^2+v^2+w^2$
$\Rightarrow u^2+v^2+w^2=u^2-6u+9+v^2+u^2$
$6u=9$
$\Rightarrow u=\frac{3}{2}$
(ii) $P{O}^2=P{B}^2$
$\Rightarrow u^2+v^2+w^2$
$=u^2+(v-6{)}^2+w^2$
$\Rightarrow u^2+v^2+w^2=u^2+v^2-12v+36+w^2$
$v=3$
(iii) $P{O}^2=P{C}^2$
$\Rightarrow u^2+v^2+w^2=u^2+v^2+(w-9{)}^2$
$\Rightarrow u^2+v^2+w^2=u^2+v^2+w^2-18w+81$
$\Rightarrow w=\frac{81}{18}$
$\Rightarrow w=\frac{9}{2}$
Now, $\left(\frac{3}{2},3,\frac{9}{2}\right)$

$\Rightarrow \left(\frac{3(2)+2\left(\frac{3}{2}\right)}{3+2},\frac{3(5)+2(3)}{3+2},\frac{3(8)+2\left(\frac{9}{2}\right)}{3+2}\right)$
$\Rightarrow \frac{6+3}{5},\frac{15+6}{5},\frac{24+9}{5}$
$=\left(\frac{9}{5},\frac{21}{5},\frac{33}{5}\right)$