The value of λ for which the lines $\frac{2-x}{3}=\frac{3-4y}{5}=\frac{z-2}{3}$ and $\frac{x-2}{-3}=\frac{2y-4}{3}=\frac{2-z}{λ}$ are perpendicular is:

$\frac{19}{8}$

The correct answer is Option (4) → $\frac{19}{8}$

A line in the symmetric form,

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

and direction vector: $d(a,b,c)$

First line:

$\frac{2-x}{3}=\frac{3-4y}{5}=\frac{z-2}{3}$

$⇒\frac{x-2}{-3}=\frac{y-\frac{3}{4}}{-\frac{5}{4}}=\frac{z-2}{3}$

$d_1=\left(-3,-\frac{5}{4},3\right)$

Second line:

$\frac{x-2}{-3}=\frac{2y-4}{3}=\frac{2-z}{λ}$

$⇒\frac{x-2}{-3}=\frac{y-2}{\frac{3}{2}}=\frac{z-2}{-λ}$

$d_2=\left(-3,\frac{3}{2},-λ\right)$

$d_1.d_2=0$

$(-3)(-3)+\frac{3}{2}×-\frac{5}{4}-3λ=0$

$⇒3λ=\frac{15}{8}-9=\frac{15-72}{8}$

$⇒λ=\frac{19}{8}$