 Solution

When $p=1$ but $t^2\ne 3t-2$, the dependent coefficient rows would require an impossible equality between their right-hand sides. Thus the augmented matrix has larger matrix rank than the coefficient matrix, and the linear system is inconsistent. For $p\ne 1$ part (i) already supplies a unique solution. Consequently there is no solution exactly when $p=1$ and $t\notin \{1,2\}$.