The exercise says that $t$ is in the plane $\pi: x-y+z =0$ and is concurrent to the lines:
$$r:\\x+y+2z=2\\x=y$$
and $$s:\\z=x+2\\y=0$$
I've transformed $r$ to the form:
$$r:\\x = \lambda\\ y = \lambda\\ z = 1-\lambda$$
and $$s:\\x = t\\y = 0\\z = t+2$$
Then the vector $$\vec {PQ}$$ from line $r$ to $s$ is:
$$(t-\lambda, -\lambda, t+1+\lambda)$$
Since the line is contained in the plane, $\vec {PQ}$ must be orthogonal to the normal $$\vec n = (1,-1,1)$$ and therefore:
$$(t-\lambda, -\lambda, t+1+\lambda)\cdot(1,-1,1) = 0$$
But I can't solve after this because I'm left with $t$ and $\lambda$. Please help me, I'm really trying :(
The line $t$ you are looking for
Therefore I'd check where $r$ resp. $s$ intersect $\pi$. The line connecting these points would be $t$.