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Q.
$\Delta A B C$ is formed by $A(1,8,4), B(0,-11,4)$ and $C(2,-3,1) .$ If $D$ is the foot of the perpendicular from $A$ to $B C$. Then the coordinates of $D$ are
Equation of line $B C$ is
$\frac{x}{2}=\frac{y+11}{8}=\frac{z-4}{-3}=\lambda$ (say)
Thus, coordinates of any point on the line $B C$ are
$(2 \lambda, 8 \lambda-11,-3 \lambda+4)$
$\therefore $ Coordinates of $D$ are of form
$(2 \lambda, 8 \lambda-11,-3 \lambda+4)$
Now, DR's of $B C$ are $2,8,-3$
and DR's of $A D$ are $2 \lambda-1,8 \lambda-19,-3 \lambda$
$\because A D \perp B C$
$\therefore 2(2 \lambda-1)+8(8 \lambda-19)+(-3)(-3 \lambda)=0$
$\Rightarrow 4 \lambda-2+64 \lambda-152+ 9 \lambda=0$
$\Rightarrow 77 \lambda=154$
$\Rightarrow \lambda=2$
Hence, the coordinates of $D$ are $(4,5,-2)$.
