Solution:

$\begin{array}{l}
\vec{a}=2 \hat{\imath}+\lambda \hat{\jmath}+3 \hat{k} \\
\vec{b}=3 \hat{\imath}+2 \hat{\jmath}-4 \hat{k}
\end{array}$
Given $\vec{a} \perp \vec{b}$
$\begin{array}{l}
\vec{a} \cdot \vec{b}=0 \\
6+2 \lambda-12=0 \\
2 \lambda=6 \\
\lambda=3
\end{array}$