This problem can be converted into a linear algebra problem. The condition is that if the derminant below is not zero, then the system has one solution.
| k 1 1 |
| 1 k 1 | = k^3 - 3k + 2 = 0
| 1 1 k |
Solving for the roots, k = -2, and k = 1.
When k = 1, the three equations are the same so there are infinite solutions.
When k = -2, there are no solutions.
When k /= -2 and k /= 1, there is one solution.