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Suppose ${\mathbf{x}}_0$ is one solution. By linearity,Thus the solution set of a linear system is exactly ${\mathbf{x}}_0+\ker T_A$. It has one member if the kernel is $\{0\}$. Conversely, if there is a unique solution and $\mathbf{z}\in \ker T_A$, then ${\mathbf{x}}_0+\mathbf{z}$ is also a solution, forcing $\mathbf{z}=0$. Combining this with i gives

Yes. A surjective $T_A$ has $\mathrm{im}{T}_A={\mathbb{R}}^n$, so every $\mathbf{b}$ is attained. Its rank of a linear map is $n$, and the rank-nullity theorem givesThe solution set of a linear system is therefore a translate of a nonzero vector subspace, containing infinitely many solutions as in iii.

By i, a free choice of $x_3$ is possible precisely when $\lambda =5$ and $\mu =2$. ThusThe complete solution set of a linear system isThe direction vector spans the kernel, and the displayed straight line is the entire solution set of a linear system.