Step back from the algebra and count degrees of freedom: three unknowns \(a,b,c\) with only two independent linear equations (the coefficient rows \((2,-5,11)\) and \((11,10,-2)\) are not multiples of each other, so both equations are genuinely independent) leaves exactly one free parameter, meaning the solution set is a line in \((a,b,c)\)-space, not a single point. The quantity \(a^2-b^2+c^2\) is a quadratic function evaluated along that line, and it will only ever give one fixed output if that quadratic happens to collapse to a constant along the line, which does not happen here since it varies linearly with the free parameter instead of cancelling out. With one continuous degree of freedom left over and the target expression genuinely depending on it, there is no way to pin down a single numeric value from the two given equations alone, so the correct choice is \(\boxed{\text{CBD (option 4)}}\), not the value \(1\), which only appears at one special point on that line.