If the lines $\frac{1-x}{3}=\frac{y-2}{2\lambda }=\frac{z-3}{2}$ and $\frac{x-1}{3\lambda }=\frac{y-1}{1}=\frac{6-z}{5}$ are perpendicular then value of $\lambda $ is :

$-\frac{10}{7}$

The correct answer is Option (2) → $-\frac{10}{7}$

$l_1:\frac{1-x}{3}=\frac{y-2}{2\lambda }=\frac{z-3}{2}$, $l_2:\frac{x-1}{3\lambda }=\frac{y-1}{1}=\frac{6-z}{5}$

$l_1⊥l_2⇒-3×3λ+2λ×1+2×-5=0$

so $-9λ+2λ-10=0$

$-10=7λ$

$λ=-\frac{10}{7}$