The corrdinates of the point where the line $\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}$ crosses XY plane is :

$\left(\frac{13}{5},\frac{23}{5}, 0 \right)$

The correct answer is Option (2) → $\left(\frac{13}{5},\frac{23}{5}, 0 \right)$

$\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=λ$

so $x=2λ+3,y=4-3λ,z=5λ+1$

at crossing of xy plane

$z = 0⇒λ=-\frac{1}{5}$

so $x=\frac{13}{5},y=\frac{23}{5},z=0$