Finding the equation of a line in a plane, concurrent with two lines that are themselves determined by pairs of planes I need to find a vectorial line equation that is contained in the plane $\pi : x - y + z = 0$ and the line t is concurrent with the lines
$$ r : \left\{ \begin{array}{lcr} x+y+2z-2 & = & 0 \\ x-y & = & 0 \end{array} \right. \qquad s : \left\{ \begin{array}{lcr} x-z+2 & = & 0 \\ y & = & 0 \end{array} \right.$$
I found the vectorial line equation of r and s, but I don't know what to do after this.
$$\begin{align}
r : (\phantom{-}2,2,-1) + \alpha(\phantom{-}2,2,-2) \\ 
s : (-1,0,\phantom{-}1) + \beta(-1,0,-1) 
\end{align}$$
 A: Hint.
Given the lines $r,s,t$ and the plane $\pi$
$$
\cases{
r\to p=p_1 + \lambda_1\vec v_1\\
s\to p=p_2 + \lambda_2\vec v_2\\
\pi\to (p-p_0)\cdot\vec n_0=0 }
$$
we have
$$
\cases{r\cap \pi\to \lambda_1^* = \frac{(p_0-p_1)\cdot\vec n_0}{\vec v_1\cdot\vec n_0}\\
s\cap \pi\to \lambda_2^* = \frac{(p_0-p_2)\cdot\vec n_0}{\vec v_2\cdot\vec n_0}\\
t\in \pi\to p = p_1 + \lambda_1^*\vec v_1+\mu\left(p_2 + \lambda_2^*\vec v_2-p_1 - \lambda_1^*\vec v_1\right)\\
}
$$
Here
$$
\cases{
p_1 = (1,1,0),\ \
\vec v_1 = (1,1,-1)\\
p_2 = (-2,0,0),\ \
\vec v_2 = (1,0,1)\\
p_0 = (1,1,0),\ \
\vec n_0 = (1,-1,1)
}
$$

NOTE
In red $t$, in blue $s, r$, in black the intersection points and in yellow $\pi$. Attached a MATHEMATICA script to build the graphics.
r1 = {1, 1, 2};
r2 = {1, -1, 0};
s1 = {1, 0, -1};
s2 = {0, 1, 0};
p0 = {1, 1, 0};
n0 = {1, -1, 1};
v1 = Cross[r1, r2]
v2 = Cross[s1, s2]
p1xy = Solve[{x + y + 2 z - 2 == 0, x - y == 0}, {x, y}][[1]];
p1 = {x, y, z} /. p1xy /. {z -> 0}
p2xy = Solve[{x - z + 2 == 0, y == 0}, {x, y}][[1]];
p2 = {x, y, z} /. p2xy /. {z -> 0}
l1 = (p0 - p1).n0/v1.n0
l2 = (p0 - p2).n0/v2.n0
p10 = p1 + l1 v1
p20 = p2 + l2 v2
t = p10 mu + (1 - mu) p20
gr1 = ParametricPlot3D[p1 + lambda v1, {lambda, -2, 2}];
gr2 = ParametricPlot3D[p2 + lambda v2, {lambda, -2, 2}];
gr0 = ContourPlot3D[x - y + z == 0, {x, -4, 4}, {y, -4, 4}, {z, -4, 4}, Mesh -> None, ContourStyle -> Directive[Yellow, Opacity[0.6],Specularity[White, 30]]];
grt = ParametricPlot3D[t, {mu, 0, 1}, PlotStyle -> Red];
grp1 = Graphics3D[{Black, Sphere[p10, 0.1]}];
grp2 = Graphics3D[{Black, Sphere[p20, 0.1]}];
Show[gr1, gr2, gr0, grt, grp1, grp2]

A: If the line $r\subseteq\pi\equiv x-y+z=0$, then the vector that defines the direction of $r$ is perpendicular to the orthogonal vector of $\pi$. Let $dir(r)=Span( \underline u )$ be the direction of $r$ generated by the vector $\underline u$ and $\underline v=(1,-1,1)$ the orthogonal vector to $\pi\implies\langle \underline u,\underline v\rangle=0$, so we can take $\underline u=(0,1,1)$.
This means that $r$ has an equation like $\begin{cases}x=\alpha\\y=\beta+t\\z=\gamma+t \end{cases}$ and if it has to be concurrent with $h\equiv\begin{cases}x+y+2z-2=0\\x-y=0 \end{cases}$ and $k\equiv\begin{cases}x-z+2=0\\y=0 \end{cases}$, we'll have that a generic point $r \ni P=(\alpha,\beta+t,\gamma+t)$ satisfies the equation of $h$ and $k$. In conclusionn, we can find the line $r$ solving the following system:
$$\begin{cases}\alpha+\beta+t+2\gamma+2t-2=0\\\alpha-\beta-t=0\\\alpha-\gamma-t+2=0\\\gamma+t=0 \end{cases}.$$
