$x + y + z = 6, \quad x + 2y + 3z = 10, \quad x + 2y + az = \beta$
Step-by-Step Solution
Key Concept: A system has unique, no, or infinitely many solutions depending on whether the determinant is nonzero, or if zero whether the augmented matrix rank equals the coefficient matrix rank.
For the system with $|A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{vmatrix} = (a-3) \neq 0$, there is a unique solution when $a \neq 3$. When $a = 3$ and $\beta \neq 10$, there is no solution (inconsistent). When $a = 3$ and $\beta = 10$, there are infinitely many solutions. At least 2 solutions implies infinitely many solutions by the linear system theory.
Correct Answer: [A-q] [B-p, s] [C-p, r] [D-p, r]