Step 1: Use the second equation to pin down x+y directly.
The system is \(x+y+kz=1\), \(2x+2y-3=0\), and \(x+2y+2kz=k\).
From the second equation, \(2x+2y=3\), so \(x+y=\frac{3}{2}\).
Step 2: Substitute into the first equation.
\(x+y+kz=1 \Rightarrow \frac{3}{2}+kz=1 \Rightarrow kz=-\frac{1}{2}\)
This single relation decides everything: if \(k\neq 0\), it fixes \(z=-\frac{1}{2k}\); but if \(k=0\), it demands \(0=-\frac{1}{2}\), which is impossible.
Step 3: Confirm a real solution exists whenever k is not zero.
With \(z=-\frac{1}{2k}\), the third equation gives \(x+2y+2kz=k \Rightarrow x+2y-1=k \Rightarrow x+2y=k+1\).
Solving this together with \(x+y=\frac{3}{2}\) gives \(y=k-\frac{1}{2}\) and \(x=2-k\), a genuine solution for every \(k\neq 0\).
Step 4: Conclusion.
Only \(k=0\) breaks the system, so
\[ \boxed{\{0\}} \]