Any point on the line $\frac{x+3}{5}=\frac{y-1}{3}=\frac{z+4}{3}=\lambda$ can be taken as $(5 \lambda-3,2 \lambda+1,3 \lambda-4)$
for foot of perpendicular $\lambda=\frac{a\left(\alpha-x_1\right)+b\left(\beta-y_1\right)+c\left(\gamma-z_1\right)}{a^2+b^2+c^2}$
$\begin{aligned}
& \Rightarrow \lambda=\frac{5(0+3)+2(2-1)+3(3+4)}{5^2+2^2+3^2} \\
& \Rightarrow \lambda=\frac{38}{38}=1
\end{aligned}$
Hence foot of perpendicular is
$(5 \times 1-3,2 \times 1+1,3 \times 1-4) \equiv(2,3,-1)$