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If $\mathrm{P}=(3,12,4)$ and $\mathrm{Q}$ is a point on the line OP such that $O Q=3$ then the sum of all the coordinates of $Q$ is
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The correct answer is:
$\pm \frac{57}{13}$
$\begin{aligned} & \text { Given } \mathrm{P}=(3,12,4) \\ & \mathrm{O} Q=3\end{aligned}$
$\Rightarrow \mathrm{QP}=10$
Now sum of all coordinates of $Q=$
$\begin{aligned} & =\left(\frac{10 \times 0+3 \times 3}{3+10}+\frac{10 \times 0+3 \times 12}{3+10}+\frac{10 \times 0+3 \times 4}{3+10}\right) \\ & = \pm \frac{57}{13}\end{aligned}$
$\Rightarrow \mathrm{QP}=10$
Now sum of all coordinates of $Q=$
$\begin{aligned} & =\left(\frac{10 \times 0+3 \times 3}{3+10}+\frac{10 \times 0+3 \times 12}{3+10}+\frac{10 \times 0+3 \times 4}{3+10}\right) \\ & = \pm \frac{57}{13}\end{aligned}$
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