= Solution
Subtracting the third row from the first gives $z=-1$, and the third row gives $y=2-x$. The second row then reduces to
$$
(1-t)x=s-t.
$$
For $t\ne1$, the unique solution is
$$
\boxed{x=\frac{s-t}{1-t},\qquad y=\frac{2-t-s}{1-t},\qquad z=-1}.
$$
Consequently $s=1$ gives $\boxed{(x,y,z)=(1,1,-1)}$ for every $t\ne1$, whereas $s=6$ gives
$$
\boxed{(x,y,z)=\left(\frac{6-t}{1-t},\frac{-4-t}{1-t},-1\right),\qquad t\ne1}.
$$
When $t=1$, the first and second left-hand sides coincide. If $s=1$, they are the same <plane>, and the remaining independent <plane> cuts it in the line
$$
\boxed{(x,y,z)=(\lambda,2-\lambda,-1),\qquad\lambda\in\mathbb R}.
$$
If $s=6$, the two <planes> are distinct and parallel, so \b[there is no solution]. Geometrically, for $t\ne1$ the three <planes> intersect at a single point; at $t=1$ the coincident-plane case leaves an entire line, while the parallel-plane case has empty intersection.
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