= Solution
At $\mu=-1$, the first equation gives $x=y$, and the second then gives $z=0$. All nonzero solutions are
$$
\boxed{\mathbf r=t(1,1,0)^{\mathsf T},\qquad t\ne0\quad(\mu=-1).}
$$
At $\mu=6$, the first equation gives $x=-6y$. The second then gives $3z=7y$, and the third is twice the second. Choosing an integer direction <vector>, all nonzero solutions are
$$
\boxed{\mathbf r=t(-18,3,7)^{\mathsf T},\qquad t\ne0\quad(\mu=6).}
$$
Each full <null space>, including the zero solution, is a line through the origin. Geometrically, the three homogeneous constraint planes have that common intersection; in each case only two of their equations are independent.
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