We have

$P_{1}:\ \ \ {\mathbf {x}}={\begin{pmatrix}1\\1\\1\end{pmatrix}}+s_{1}{\begin{pmatrix}0\\1\\0\end{pmatrix}}+s_{2}{\begin{pmatrix}1\\2\\-1\end{pmatrix}}$

and

$P_{2}:\ \ \ {\mathbf {x}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+t_{2}{\begin{pmatrix}0\\1\\2\end{pmatrix}}$

for some $s_{1}$, $s_{2}$, $t_{1}$ and $t_{2}$ from the question and part (b).

It is easy to see that the line $l_{2}$ is just the intersection of plane $P_{1},P_{2}$, i. e., any point ${\mathbf {x}}$ on the line has to satisfy both equations. Thus we have following equation for line $l_{2}$ .

${\mathbf {x}}={\begin{pmatrix}1\\1\\1\end{pmatrix}}+s_{1}{\begin{pmatrix}0\\1\\0\end{pmatrix}}+s_{2}{\begin{pmatrix}1\\2\\-1\end{pmatrix}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+t_{2}{\begin{pmatrix}0\\1\\2\end{pmatrix}}$.

Now let's find out the relation among parameters $s_{1},s_{2},t_{1},t_{2}$ .
The first coordinate shows $1+s_{2}=-t_{1}$, while the third shows $1-s_{2}=-t_{1}+2t_{2}$. Summing these equations gives $2=-2t_{1}+2t_{2}$, hence $t_{2}=1+t_{1}$. Substituting it back gives: for any point ${\mathbf {x}}$ on line it satisfies

$l_{2}:{\mathbf {x}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+(1+t_{1}){\begin{pmatrix}0\\1\\2\end{pmatrix}}=t_{1}{\begin{pmatrix}-1\\1\\1\end{pmatrix}}+{\begin{pmatrix}0\\1\\2\end{pmatrix}},$

which has is equivalent equation form; for any point ${\mathbf {x}}=(x_{1},x_{2},x_{3})^{T}$ on the line $l_{2}$ satisfies

$\color {blue}{\begin{aligned}{\begin{cases}x_{1}+x_{2}=1\\x_{1}+x_{3}=2\end{cases}}\end{aligned}}$