Equation of line through $\mathrm{B}(0,4,1)$ and $\mathrm{C}(-2,0,4)$
$$
\begin{aligned}
& \left.\right|_{\mathrm{D}} ^{\mathrm{A}(2,0,3)} \\
& \frac{x-0}{-2-0}=\frac{y-4}{0-4}=\frac{z-1}{4-1} \\
& \frac{-x}{2}=\frac{y-4}{-4}=\frac{z-1}{3}=\lambda \\
& x=-2 \lambda, y-4=-4 \lambda, \frac{z-1}{3}=\lambda \\
& y=4-4 \lambda, z=3 \lambda+1 \\
& \text { So, } \mathrm{D}(-2 \lambda, 4-4 \lambda, 3 \lambda+1) \\
& \text { Direction ratio's of } \mathrm{AD}=(-2 \lambda-2,4-4 \lambda-0,3 \lambda+1-3) \text {. } \\
& =(-2 \lambda-2,4-4 \lambda, 3 \lambda-2) \\
& \text { Direction ratio's of } \mathrm{BC}=(-2-0,0-4,4-1) \\
& =(-2,-4,3) \\
& \text { Here, } \mathrm{AD} \perp \mathrm{BC} \\
& \text { Then, }(-2)(-2 \lambda-2)-4(4-4 \lambda)+3(3 \lambda-2)=0 \\
& 4 \lambda+4-16+16 \lambda+9 \lambda-6=0 \\
& 29 \lambda-18=0 \\
& \lambda=\frac{18}{29} \\
& \text { Point D }=\left(-2 \times \frac{18}{29}, 4-4 \times \frac{18}{29}, 3 \times \frac{18}{29}+1\right) \\
& \mathrm{D}=\left(\frac{-36}{29}, \frac{44}{29}, \frac{83}{29}\right) \\
&
\end{aligned}
$$
Now, Apply section formula and suppose the ratio is $m: n$.
$$
\begin{aligned}
& \frac{-36}{29}=\frac{m \times-2+n \times 0}{m+n} \\
& -36 m-36 n=-58 m \\
& 36 n=22 m \\
& \frac{m}{n}=\frac{36}{22}=\frac{18}{11}
\end{aligned}
$$