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 $\displaystyle \alpha$, $\displaystyle \beta$, $\displaystyle \gamma$there is always a solution if

[a b c] = $\displaystyle \alpha$[3 0 2] + $\displaystyle \beta$[4 1 3] + $\displaystyle \gamma$[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 $\displaystyle \left(\begin{array}{c}1 \\0 \\0\end{array}\right)$, $\displaystyle \left(\begin{array}{c}0 \\1 \\0\end{array}\right)$, and $\displaystyle \left(\begin{array}{c}-5 \\4 \\0\end{array}\right)$.

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