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Let $A, B, C$ be three points on $\overline{O X}, \overline{O Y}, \overline{O Z}$ 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 $P Q$ in the ratio $3: 2$ is
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Verified Answer
The correct answer is:
$\left(\frac{9}{5}, \frac{21}{5}, \frac{33}{5}\right)$
Let $P$ is $(u, v, w)$.
Here, $P O^2=P A^2, P O^2=P B^2, P O^2=P C^2$
$$
\begin{aligned}
u^2+v^2+w^2 & =(u-3)^2+v^2+w^2 \\
u^2+v^2+w^2 & =u^2-6 u+9+v^2+u^2 \\
6 u & =9 \Rightarrow u=\frac{3}{2}
\end{aligned}
$$
(ii)
$$
\begin{aligned}
\mathrm{PO}^2=P B^2 \Rightarrow u^2 & +v^2+w^2 \\
& =u^2+(v-6)^2+w^2 \\
\Rightarrow \quad u^2+v^2+w^2 & =u^2+v^2-12 v+36+w^2 \\
v & =3
\end{aligned}
$$
$$
\begin{array}{lrl}
\text { (iii) } & P O^2=P C^2 \\
\Rightarrow & u^2+v^2+w^2 & =u^2+v^2+(w-9)^2 \\
\Rightarrow & u^2+v^2+w^2 & =u^2+v^2+w^2-18 w+81 \\
\Rightarrow & w & =\frac{81}{18} \Rightarrow w=\frac{9}{2}
\end{array}
$$
Now,
$$
\begin{aligned}
& \Rightarrow\left(\frac{3(2)+2\left(\frac{3}{2}\right)}{3+2}, \frac{3(5)+2(3)}{3+2}, \frac{3(8)+2\left(\frac{9}{2}\right)}{3+2}\right) \\
& \Rightarrow \frac{6+3}{5}, \frac{15+6}{5}, \frac{24+9}{5}=\left(\frac{9}{5}, \frac{21}{5}, \frac{33}{5}\right)
\end{aligned}
$$
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