The point of intersection of the xy plane and the line passing through the points A ≡ (3, 4, 1) and B ≡ (5, 1, 6) is :

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

Direction ratios of AB are 2, -3, 5

Thus equation of AB is

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

For the point of intersection of this line with xy–plane, we have

Z = 0

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

$\Rightarrow x=3-\frac{2}{5}=\frac{13}{5}, y=4+\frac{3}{5}=\frac{23}{5}$

Hence, the required point is $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$