The system

3x+4y+z=a

y+4z=b

2x+3y+2z=c

has an obvious solution if [a b c] = [3 0 2]or[4 1 3]or[1 4 2]

find these solutions.

Prove or disprove: for any numbers , , there is always a solution if

[a b c] = [3 0 2] + [4 1 3] + [1 4 2]

Under what gerenal conditions on a,b,c will there always be a solution?

the matrix [a b c] [3 0 2] [4 1 3] [1 4 2]supposed to be in column. I am kind of confused with the question

I have solved the a part the 3 solution is , , and .

now how do I prove there is always a solution under what condition