Since equation of line joining points \(B(4,7,1)\) and \(C(3,5,3)\) is
\(\begin{aligned}
\frac{x-4}{4-3} & =\frac{y-7}{7-5}=\frac{z-1}{1-3} \\
\Rightarrow \quad \frac{x-4}{1} & =\frac{y-7}{2}=\frac{3-1}{-2}=\lambda \quad \text{(let)} ...(i)
\end{aligned}\)
Now the general point over the above line (i) is \(P(\lambda+4,2 \lambda+7,1-2 \lambda)\)
Let point \(P\) is the foot of perpendicular of point \(A(1,0,3)\) over the line (i), so
AP \(\perp\) line
\(\begin{aligned}
& \Rightarrow(\lambda+4-1)(1)+(2 \lambda+7-0)(2)+(1-2 \lambda-3)(-2)=0 \\
& \Rightarrow \lambda+3+4 \lambda+14+4 \lambda+4=0 \Rightarrow 9 \lambda+21=0 \\
& \Rightarrow \quad \lambda=-\frac{7}{3} \\
\end{aligned}\)
\(\therefore\) The coordinate of point \(P\) is
\(\left(4-\frac{7}{3}, 7-\frac{14}{3}, 1+\frac{14}{3}\right)=\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)\)
Hence, option (a) is correct.