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

$\Rightarrow \, \left(\frac{3(2)+2\left(\frac{3}{2}\right)}{3+2}, \frac{3(5)+2(3)}{3+2}, \frac{3(8)+2\left(\frac{9}{2}\right)}{3+2}\right)$
$\Rightarrow \,\frac{6+3}{5}, \frac{ 1 5+6}{5}, \frac{24+9}{5}$
$=\left(\frac{9}{5}, \frac{21}{5}, \frac{33}{5}\right)$