Given as 2x – y + 2z = 0
5x + 3y – z = 0
X + 5y – 5z = 0
The equation can be written as

AX = 0
Then |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)
|A| = – 20 – 24 + 44
|A| = 0
So, the system has infinite solutions
Suppose z = k
2x – y = – 2k
5x + 3y = k

So, x = -5k/11, y = 12k/11 and z = k