The value on. for which the lines $\frac{x-1}{2}=\frac{y}{3}=\frac{z+1}{4}$ and $\frac{x+4}{3}=\frac{y-z}{\lambda }=\frac{z-5}{3}$

are perpendicular is:

Two lines 

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$

are perpendicular, iff $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

Thus, in this question

$\begin{array}{l}2\times 3+3\times \lambda +4\times 3=0\\ \Rightarrow \lambda =-6\end{array}$