Matrix solution to solve c1, c2, c3

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Task: Determine which of the given numbers is the second coordinate of the vector x=[4; 5; 0] in the base {b1, b2, b3} where b1=[2; 1; 3], b2=[3; 1; 4] and b3=[-1; 3; 1]

Given numbers: 2, -1, -3 or 1

My answer: I guess that I should use c1*b1+c2*b2+c3*b3=x and in that way calculate each value of c1, c2 and c3.

c1*[2; 1; 3]+c2*[3; 1; 4]+c3*[-1; 3; 1]=[4; 5; 0]

2*c1+3*c2-c3=4

c1+c2+3*c3=5

3*c1+4*c2+c3=0

I get it to:

3c1-4*c2+c3=0

0+17c2+c3=12

0+0+43*c3=57

How do i do?

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Hint: Multiplying the second equation by $(-2)$ and adding to the first $$c_2-7c_3=-6$$ multiplying the second equation by $-3$ and adding to the third then we get $$c_2-8c_3=-15$$ and from here we obtain $$-c_3=-9$$ Can you finish now?