The system of linear equations \[\left[ {\begin{array}{*{20}{c}} 2&1&3 \\ 3&0&1 \\ 1&2&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\text{a}} \\ {\text{b}} \\ {\text{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ { – 4} \\ {14} \end{array}} \right]\] has A. a unique solution B. infinitely many solutions C. no solution D. exactly two solutions

The correct answer is $\boxed{\text{A}}$.

To solve a system of linear equations, we can use Gaussian elimination. In this case, we can reduce the system to the following form:

$$\left[ {\begin{array}{{20}{c}} 1&0&-2 \ 0&1&-3 \ 0&0&1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\text{a}} \ {\text{b}} \ {\text{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \ { – 4} \ {14} \end{array}} \right]$$

This means that the system has a unique solution, which is $\boxed{\text{a} = 5, \text{b} = -4, \text{c} = 14}$.

Here is a brief explanation of each option:

• Option $\boxed{\text{A}}$: The system has a unique solution. This is because the reduced matrix is invertible, which means that there is a unique solution to the system of equations.
• Option $\boxed{\text{B}}$: The system has infinitely many solutions. This is because the reduced matrix is not invertible, which means that there are infinitely many solutions to the system of equations.
• Option $\boxed{\text{C}}$: The system has no solution. This is because the reduced matrix is singular, which means that there is no solution to the system of equations.
• Option $\boxed{\text{D}}$: The system has exactly two solutions. This is not possible, because the system either has a unique solution, infinitely many solutions, or no solution.