The points $\left(3, -2, \mathrm{-\lambda }\right)$ lie on Plane $2x-4y+3z-2=0$

$\Rightarrow 6+8-3\lambda -2=0$

$\Rightarrow 3\lambda =12$

$\Rightarrow \lambda =4$

Now,

$\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z+4}{-2}$.......$\left(i\right)$

$\frac{x-1}{12}=\frac{y}{9}=\frac{z}{4}$........$\left(ii\right)$

Shortest distance formula

$d=\left|\frac{\left|\begin{array}{ccc}{x}_1-x_2& y_2-y_1& z_2-z_1\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|}{\left|\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|\right|}\right|$

Calculating we get the value as $0$
Hence lines are intersecting (As $\overrightarrow{b}\nparallel \overrightarrow{d}$ )

$\therefore$ Shortest distance$=0$