The point of intersection of the lines and is:

(d) $(1, 2, 3)$ The equations of the lines are $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-1}{5} = \frac{y-2}{2} = \frac{z-3}{1}$. Substituting the point $(1, 2, 3)$ into both lines: For line 1: $\frac{1-1}{2} = \frac{2-2}{3} = \frac{3-3}{4} = 0$. For line 2: $\frac{1-1}{5} = \frac{2-2}{2} = \frac{3-3}{1} = 0$. Both lines pass through $(1, 2, 3)$, so their point of intersection is $(1, 2, 3)$.