Given as 3x – y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0
The equation can be written as

AX = 0
Then, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)
|A| = – 27 + 1 + 26
|A| = 0
So, the system has infinite solutions
Suppose z = k
3x – y = – 2k
4x + 3y = – 3k

So, x = (-9k/13), y = (-k/13) and z = k