Yeah you can do it via back substitution (or just inspection). As usual, the matrix U represents linear equations in the variables alpha, beta, gamma. (First column is for coefficients of alpha, second for beta, and third for gamma.) So they're asking for a non-zero solution to the system of equations described by that matrix.
Oh, and it only asks us to give a specific such solution (which is why I said it could be done via inspection). If you use back-substitution, you don't need to worry about capturing all the solutions for the purposes of this question, just need to give one particular one.
This means if you're doing back-substitution and are about to set a non-leading variable (just gamma here) to a parameter, you can just set it to a particular number (like 1, or 5, or whatever specific non-zero number you like; choosing 5 is one way to make the arithmetic easy, as it'll mean you avoid fractions popping up when solving for the other variables. If you picked 0 as the number, all the other variables would come out to 0 too, which isn't what we want.). Then you can solve for numerical values of the other variables, which will give you a desired solution.
 = \cos t$ and $\frac{\mathrm{d}}{\mathrm{d}t}\left(\cos t \right) = -\sin t$. Also make sure to recall the chain rule. $)
