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

• A

  $\pm \frac{10}{13}$
• B

  $\pm \frac{28}{13}$
• C

  $\pm \frac{19}{13}$
• Correct answer
  D

  $\pm \frac{57}{13}$

Solution

$\begin{aligned} & \text { Given } \mathrm{P}=(3,12,4) \\ & \mathrm{O} Q=3\end{aligned}$ $\Rightarrow \mathrm{QP}=10$
Now sum of all coordinates of $Q=$ $\begin{aligned} & =\left(\frac{10 \times 0+3 \times 3}{3+10}+\frac{10 \times 0+3 \times 12}{3+10}+\frac{10 \times 0+3 \times 4}{3+10}\right) \\ & = \pm \frac{57}{13}\end{aligned}$

Hence, the correct answer is option: 'D'.