The system can be written as

A X = 0
Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)
|A| = – 20 – 24 + 44
|A| = 0
Hence, the system has infinite solutions
Let z = k
2x – y = – 2k
5x + 3y = k

Hence, x = \(\frac{-5k}{11},\) y = \(\frac{12k}{11}\) and z = k