Solution:

It is known to us that the vector eq. of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given as
$\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$
Therefore the position vector of point $\mathrm{A}(5,1,6)$ is given as
$\vec{a}=5 \hat{i}+\hat{j}+6 \hat{k} \ldots$
And the position vector of point $\mathrm{B}(3,4,1)$ is given as $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ldots$
$\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}} \ldots(2)$
Therefore subtract eq.(2) and eq.(1) we obtain
$\begin{aligned}
(\vec{b}-\vec{a}) =(3 \hat{i}+4 \hat{j}+\hat{k})-(5 \hat{i}+\hat{j}+6 \hat{k}) \\
=(3-5) \hat{i}+(4-1) \hat{j}+(1-6) \hat{k} \\
=(-2 \hat{i}+3 \hat{j}-5 \hat{k}) \\
\vec{r}=(5 \hat{i}+&\hat{j}+6 \hat{k})+\lambda(-2 \hat{i}+3 \hat{j}-5 \hat{k}) \\
\ldots .(3)
\end{aligned}$

Let the coordinates of the point where the line crosses the YZ plane be $(0, \mathrm{y}, \mathrm{z})$ So,
$\vec{r}=(0 \hat{\boldsymbol{i}}+\boldsymbol{\boldsymbol { j }}+\boldsymbol{z} \hat{\boldsymbol{k}}) \ldots \ldots(4)$
As the point lies in line, it satisfies its eq.,

Substituting eq.(4) in eq.(3) we obtain,
$\begin{aligned}
(0 \hat{\boldsymbol{i}}+\boldsymbol{y} \hat{\boldsymbol{j}}+\boldsymbol{z} \hat{\boldsymbol{k}}) &=(5 \hat{\boldsymbol{i}}+\hat{\boldsymbol{j}}+6 \hat{\boldsymbol{k}})+\lambda(-2 \hat{\boldsymbol{i}}+3 \hat{\boldsymbol{j}}-5 \hat{\boldsymbol{k}}) \\
=(5-2 \lambda) \hat{\boldsymbol{i}}+(1+3 \boldsymbol{\lambda}) \hat{\boldsymbol{j}}+(6-5 \lambda) \hat{\boldsymbol{k}}
\end{aligned}$
It is known to us that, two vectors are equal if their corresponding components are equal

Therefore,
$0=5-2 \lambda$ $5=2 \lambda$ $\lambda=5 / 2$ $\mathrm{y}=1+3 \lambda$ And,
$z=6-5 \lambda \ldots(6)$

Substituting the value of $\lambda$ in eq. $(5)$ and $(6)$, we obtain $-$
$\begin{array}{l}
y=1+3 \lambda \\
=1+3 \times(5 / 2) \\
=1+(15 / 2)
\end{array}$
$=17 / 2$

And
$\begin{array}{l}
z=6-5 \lambda \\
=6-5 \times(5 / 2) \\
=6-(25 / 2) \\
=-13 / 2
\end{array}$

As a result, the coordinates of the required point is $(0,17 / 2,-13 / 2)$.