This is the same as showing the following system of equations doesn't have a solution:
or in matrix form,
The quickest way to check if there is a solution is to check whether the coefficient matrix is invertible. If its determinant is 0, then it is not invertible.
And the quickest way to show that the determinant is 0 is by observing that the third row is a linear combination of the first two rows:
(-2, 9, 6) - (-3, 2, 1) = (-2 + 3, 9 - 2, 6 - 1) = (1, 7, 5)
So there are indeed no such scalars <em>c₁</em>, <em>c₂</em>, and <em>c₃</em>.