Answer: The coefficient matrix is singular when

\(\begin{vmatrix}k&1&1\\1&k&1\\1&1&k\end{vmatrix}=0.\)

This determinant is \(k^3-3k+2\), so

\(k^3-3k+2=(k-1)^2(k+2)=0.\)

Therefore \(k=1\) or \(k=-2\).

For \(k=1\), the equations are inconsistent, so there is no solution.

For \(k=-2\), the solutions are \((x,y,z)=(t-1,t,t)\), where \(t\in\mathbb R\).

Let

\(A=\begin{pmatrix}k&1&1\\1&k&1\\1&1&k\end{pmatrix}.\)

For the system to have no unique solution, we need \(\det A=0\).

Now

\(\det A=\begin{vmatrix}k&1&1\\1&k&1\\1&1&k\end{vmatrix}=k^3-3k+2.\)

So

\(k^3-3k+2=(k-1)^2(k+2)=0.\)

Hence \(k=1\) or \(k=-2\).

When \(k=1\), the equations become

\(x+y+z=2,\qquad x+y+z=-1,\qquad x+y+z=-1.\)

These cannot all be true at the same time, so the system has no solution.

When \(k=-2\), the system is

\(-2x+y+z=2,\qquad x-2y+z=-1,\qquad x+y-2z=-1.\)

Subtracting the second equation from the third gives

\(3y-3z=0,\)

so \(y=z\).

Substitute \(y=z\) into \(x-2y+z=-1\):

\(x-z=-1,\)

so \(x=z-1\). Let \(z=t\). Then

\((x,y,z)=(t-1,t,t),\qquad t\in\mathbb R.\)