The value of $k$ for which the lines $\frac{2x-3}{4}=\frac{3-y}{k}=\frac{z-2}{-2}$ and $\frac{x-2}{1}=\frac{y}{4}=\frac{5-z}{3}$ are perpendicular to each other is:

2

The correct answer is Option (3) → 2

$\frac{2x-3}{4}=\frac{3-y}{k}=\frac{z-2}{-2}\quad\Rightarrow\quad (a_1,b_1,c_1)=(2,-k,-2)$

$\frac{x-2}{1}=\frac{y}{4}=\frac{5-z}{3}\quad\Rightarrow\quad (a_2,b_2,c_2)=(1,4,-3)$

Perpendicular condition: $2\cdot1+(-k)\cdot4+(-2)\cdot(-3)=0$

$2-4k+6=0$

$k=2$