Answer: The system has a unique solution when the determinant of the coefficient matrix is non-zero.

For \(\begin{pmatrix}1&2&3\\k&4&6\\7&8&9\end{pmatrix}\),

\(\det = 6k-12 = 6(k-2)\).

So a unique solution exists when \(6(k-2)\neq 0\), hence \(k\neq 2\).

A system of three linear equations has a unique solution exactly when the coefficient matrix is invertible, so its determinant must be non-zero.

The coefficient matrix is

\(A=\begin{pmatrix}1&2&3\\k&4&6\\7&8&9\end{pmatrix}\).

Expand the determinant along the first row:

\(\det(A)=1\begin{vmatrix}4&6\\8&9\end{vmatrix}-2\begin{vmatrix}k&6\\7&9\end{vmatrix}+3\begin{vmatrix}k&4\\7&8\end{vmatrix}.\)

Now evaluate the \(2\times 2\) determinants:

\(\begin{vmatrix}4&6\\8&9\end{vmatrix}=36-48=-12,\)

\(\begin{vmatrix}k&6\\7&9\end{vmatrix}=9k-42,\)

\(\begin{vmatrix}k&4\\7&8\end{vmatrix}=8k-28.\)

So

\(\det(A)=-12-2(9k-42)+3(8k-28).\)

Simplifying:

\(\det(A)=-12-18k+84+24k-84=6k-12=6(k-2).\)

For a unique solution,

\(\det(A)\neq 0.\)

Hence

\(6(k-2)\neq 0 \Rightarrow k\neq 2.\)

Therefore, the set of values of \(k\) is \(\{k\in\mathbb{R}:k\neq 2\}.\)