Let foot of perpendicular be $\mathrm{B}\equiv \left(-2,\frac{7}{2},\frac{3}{2}\right)$
D.R.'s of line A B are$<-1+2,4-\frac{7}{2},3-\frac{3}{2}>$
$\mathrm{i}.\mathrm{e}.,<2,1,3>$
$\mathrm{AB} \mathrm{is} \perp \mathrm{r} \mathrm{to} \mathrm{plane}, \mathrm{so} \mathrm{D}.\mathrm{R}.\text{'}\mathrm{s}$of normal of plane will be $<2,1,3>$
$\therefore \mathrm{m}=1,\mathrm{n}=3$
Now, equation of AC is
$\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}-4}{-1}=\frac{\mathrm{z}-3}{-4}=\mathrm{\lambda }\left(\text{ say }\right)$
Any point that is C on AC is $(3\mathrm{\lambda }-1,-\mathrm{\lambda }+4,-4\mathrm{\lambda }+3)$
C lies on plane $2\mathrm{x}+\mathrm{y}+3\mathrm{z}=4$
$\begin{array}{l}\therefore 6\mathrm{\lambda }-2-\mathrm{\lambda }+4-12\mathrm{\lambda }+9=4\Rightarrow -7\mathrm{\lambda }=-7\Rightarrow \mathrm{\lambda }=1\\ \therefore \mathrm{C}\equiv (2,3,-1)\end{array}$

Thus, $\mathrm{AC}=\sqrt{(2+1{)}^2+(3-4{)}^2+(-1-3{)}^2}$
$=\sqrt{9+1+16}=\sqrt{26} units$