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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.

The elimination in i determines $x_3$ uniquely exactly when $\lambda \ne 5$. The other two coordinates are then also fixed. Explicitly,For $\lambda =5$, the reduced equation is either inconsistent or leaves $x_3$ free, so a unique solution is impossible.