= Solution
The coefficient <matrix> has <determinant>
$$
\det\begin{pmatrix}p&1&1\\1&2&4\\1&4&10\end{pmatrix}
=4p-6+2=4(p-1).
$$
A square <linear system> has a unique solution for every right-hand side precisely when its coefficient <matrix> is an <invertible matrix>. Here \b[a unique solution exists exactly when $p\ne1$, for every real $t$].
For an explicit <Gaussian elimination>, subtract twice the second equation from the third, then use the second equation, obtaining $x=2z+2t-t^2$ and $y=(t^2-t)/2-3z$. The first equation determines
$$
z=\frac{2+(2p-1)t^2+(1-4p)t}{4(p-1)},
$$
and the preceding formulas determine $x,y$. When $p=1$ this last coefficient vanishes; the next parts determine compatibility rather than dividing by zero.
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