Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane
It is known that the equation of the line passing through the points, $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$, is $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,
$\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$
$\Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k$ (say)
$\Rightarrow x=5-2 k, y=3 k+1, z=6-5 k$
Any point on the line is of the form $(5-2 k, 3 k+1,6-5 k)$.
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane,
$5-2 k=0$
$\Rightarrow k=\frac{5}{2}$
$\Rightarrow 3 k+1=3 \times \frac{5}{2}+1=\frac{17}{2}$
$6-5 k=6-5 \times \frac{5}{2}=\frac{-13}{2}$
Therefore, the required point is $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$.