The equations of line through (5, 1, 6) and (3, 4, 1) are$\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$$\Rightarrow\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}$... (1)Any point on (1) is (5 – 2k, 1 + 3k, 6 – 5k) ...(2)This lies on YZ-plane $\therefore x = 0$$\Rightarrow 5-2k=0 \Rightarrow k=\frac{5}{2}$Putting the value of $k=\frac{5}{2}$ in (2), $\left(5-5,1+\frac{15}{2},6-\frac{25}{2}\right)$i.e., $\left(0,\frac{17}{2},-\frac{13}{2}\right)$ is the required point.