The system can be written as

A X = 0
Now, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)
|A| = – 27 + 1 + 26 |A| = 0
Hence, the system has infinite solutions
Let z = k
3x – y = – 2k
4x + 3y = – 3k

Hence, x = \(\frac{-9k}{13},\) y = \(\frac{-k}{13}\) and z = k