The vertices of $\triangle A B C$ are given as $A(1,8,4)$, $B(0,-11,4)$ and $C(2,-3,1)$.
Equation of line $B C$,
$$
\frac{x}{2}=\frac{y+11}{8}=\frac{z-4}{-3}=\lambda \text { (say) }
$$


As point $D$ is on line $B C$, so coordinates of $D$ are $(2 \lambda, 8 \lambda-11,-3 \lambda+4)$
Since, $A D \perp B C$
$$
\begin{aligned}
\therefore \quad & A D \cdot B C=0 \\
& (2 \lambda-1,8 \lambda-19,-3 \lambda) \cdot(2,8,-3)=0 \\
\therefore \quad & 2(2 \lambda-1)+8(8 \lambda-19)+(-3)(-3 \lambda)=0 \\
& 4 \lambda-2+64 \lambda-152+9 \lambda=0 \\
& 77 \lambda=154 \\
& \lambda=2
\end{aligned}
$$
Hence, the coordinates of $D$ are $(4,5,-2)$