The existence of the unique solution of the system of equations

$x + y + z =λ$

$5x-y+μz=10$

$2x + 3y-z=6$

depends on

μ only

The given system of equations may be written as

$\begin{bmatrix}1&1&1\\5&-1&μ\\2&3&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}λ &10&6\end{bmatrix}$

$⇒\begin{bmatrix}1&1&1\\0&-6&μ-5\\0&1&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}λ &10-5λ&6-2λ\end{bmatrix}$  Applying $R_2 → R_2-5R_1,R_3 → R_3 -2R_1$

$⇒\begin{bmatrix}1&1&1\\0&-6&μ-5\\0&6&-18\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}λ &10-5λ&36-12λ\end{bmatrix}$ Applying $R_3→ 6R_3$

$⇒\begin{bmatrix}1&1&1\\0&-6&μ-5\\0&0&μ-23\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}λ &10-5λ&46-17λ\end{bmatrix}$ Applying $R3→ R_3 + R_2$

For unique solution we must have $μ-23≠0$. i.e. $μ ≠23$.

Hence, the existence of the unique solution depends on μ only.