The equation of line through (5, 1, 6) and (3, 4, 1) is $\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$$\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}$...(1)Any point on (1) is (5 – 2k, 1 + 3k, 6 – 5k) ...(2)This lies on ZX plane∴ y = 0$\Rightarrow 1+3k=0\Rightarrow k=-\frac{1}{3}$Putting the value of $k=-\frac{1}{3}$ in (2), we get, $\left(5+\frac{2}{3},1-1,6+\frac{5}{3}\right)$ i.e., $\left(\frac{17}{3},0,\frac{23}{3}\right),$ is the required point.