$$
\text { Let } A=(9,13,15)
$$




$$
\begin{aligned}
& B=(12,21,10) \\
& P=(5,7,3) \text { and } Q=(x, y, z)
\end{aligned}
$$
Dr's of $\overleftrightarrow{A B}=(9-12,13-21,15-10)$
$$
=(-3,-8,5)=(3,8,-5)
$$

Equation of line $\overleftrightarrow{A B}$ is
$$
\begin{aligned}
\frac{x-9}{3} & =\frac{y-13}{8}=\frac{z-15}{-5}=\lambda \\
\frac{x-9}{3} & =\lambda, \frac{y-13}{8}=\lambda, \frac{z-15}{-5}=\lambda
\end{aligned}
$$
$$
\begin{aligned}
& x=3 \lambda+9, y=8 \lambda+13, z=-5 \lambda+15 \\
& \therefore Q=(3 \lambda+9,8 \lambda+13,-5 \lambda+15)
\end{aligned}
$$

Dr's of $\mathbf{P Q}=(3 \lambda+9-5,8 \lambda+13-7,-5 \lambda+15-3)$
Dr's $^{\prime}$ of $\mathbf{P Q}=(3 \lambda+4,8 \lambda+6,-5 \lambda+12)$
Since, $\overleftrightarrow{A B} \perp \mathbf{P Q}$
$$
\begin{aligned}
a_1 a_2+b_1 b_2+c_1 c_2 & =0 \\
3(3 \lambda+4)+8(8 \lambda+6)-5(-5 \lambda+12) & =0 \\
9 \lambda+12+64 \lambda+48+25 \lambda-60 & =0 \quad \Rightarrow \lambda=0
\end{aligned}
$$
Put, $\lambda=0$ in $Q$
$$
\therefore \quad Q=(9,13,15)
$$
$\therefore$ Foot of Perpendicular $Q=(9,13,15)$
Hence, option (4) is correct.