= Solution
For $p=1$, the third coefficient row equals three times the second minus twice the first. The first two rows are independent, so the coefficient <matrix rank> is two. The same dependence must hold for the right-hand side, giving
$$
t^2=3t-2\quad\Longleftrightarrow\quad(t-1)(t-2)=0.
$$
Thus \b[infinitely many solutions occur exactly for $p=1$ and $t\in\{1,2\}$]. The <null space> has dimension one by the <rank-nullity theorem>, and an explicit parameterization, with $s\in\mathbb R$, is
$$
\boxed{\begin{array}{ll}
t=1:&(x,y,z)=(1+2s,-3s,s),\\
t=2:&(x,y,z)=(2s,1-3s,s).
\end{array}}
$$
Substitution verifies all three equations in each family.
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