Given vertices of triangle $\mathrm{A}(2,3,-1), \mathrm{B}(4,1,0)$ and $\mathrm{C}(-1,-1,11)$
$$
\begin{aligned}
& \mathrm{AB}=\sqrt{(4-2)^2+(1-3)^2+(0+1)^2} \\
& =\sqrt{4+4+1}=3 \\
& \mathrm{AC}=\sqrt{(2+1)^2+(3+1)^2+(-1-11)^2} \\
& =\sqrt{9+16+144}=\sqrt{169}=13 \\
& \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{13}
\end{aligned}
$$
Apply section formula,
$$
\begin{aligned}
& \mathrm{D}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{3 \times-1+13 \times 4}{3+13}, \frac{3 \times-1+13 \times 1}{3+13}, \quad=\left(\frac{49}{16}, \frac{10}{16}, \frac{33}{16}\right)\right. \\
& \left.\frac{3 \times 11+13 \times 0}{3+13}\right) \\
& \text { Direction ratio of } \mathrm{AD}=\left(\frac{49}{16}-2, \frac{10}{16}-3, \frac{33}{16}+1\right) \\
& =\left(\frac{17}{16}, \frac{-38}{16}, \frac{49}{16}\right)
\end{aligned}
$$
Then, direction ratio is the multiple of any constant $1 / 16$. So, direction ratio of AD would be $(17,-38,4 a)$. So, option (c) is correct.