A linear system with a parameter

Determinant. Let $A$ be the coefficient matrix:
$$A=\begin{pmatrix}1&1&1\\1&2&3\\1&3&m\end{pmatrix}.$$
Subtract row $1$ from rows $2$ and $3$:
$$\det A=\begin{vmatrix}1&1&1\\0&1&2\\0&2&m-1\end{vmatrix}=1\cdot\bigl(1\cdot(m-1)-2\cdot2\bigr)=m-5.$$

Case $m\neq5$: unique solution. Subtract the first equation from the second and from the third:
$$y+2z=3,\qquad 2y+(m-1)z=6.$$
From the first, $y=3-2z$. Substituting into the second:
$$2(3-2z)+(m-1)z=6\ \Longrightarrow\ (m-5)z=0.$$
Since $m\neq5$, this gives $z=0$, then $y=3$, and from the first equation $x=3-y-z=0$. The unique solution is
$$(x,y,z)=(0,3,0).$$

Case $m=5$: infinitely many solutions. The third equation becomes $x+3y+5z=9$. Subtracting consecutive equations:
$$L_2-L_1:\ y+2z=3,\qquad L_3-L_2:\ y+2z=3.$$
So the third equation is a consequence of the first two, and the system reduces to
$$\begin{cases}x+y+z=3\\ y+2z=3.\end{cases}$$
Choose $z=t\in\mathbb{R}$. Then $y=3-2t$ and
$$x=3-y-z=3-(3-2t)-t=t.$$
The solutions are $(x,y,z)=(t,\,3-2t,\,t)$, $t\in\mathbb{R}$.

Verification. At $t=0$: $(0,3,0)$ satisfies $0+3+0=3$, $0+6+0=6$, $0+9+0=9$. At $t=1$: $(1,1,1)$ satisfies $1+1+1=3$, $1+2+3=6$, $1+3+5=9$. The $m\neq5$ solution is the member $t=0$ of the $m=5$ family.

Conclusion.
- If $m\neq5$: unique solution $(x,y,z)=(0,3,0)$.
- If $m=5$: infinitely many solutions $(x,y,z)=(t,\,3-2t,\,t)$, $t\in\mathbb{R}$.