if [MATH]a \neq 5[/MATH](3) yields that [MATH]x_3=0[/MATH] hence:
(1) = [MATH]5x_1+4x_2+3x_3=1[/MATH]<---Not correct (unless a=5!)
(2) = [MATH]2x_2+x_3=1[/MATH]<---Not correct (unless a=5!)
and following the same method as JeffM displayed above we can solve for [MATH]x_1[/MATH] and [MATH]x_2[/MATH] quite easily because (2) gives us [MATH]x_2[/MATH] instantly and then it is all a matter of plugging [MATH]x_2[/MATH] into (1) and we are done.
Suppose a = 0:
(3) just like before instantly yields that [MATH]x_3[/MATH] must be 0. (1) we get that [MATH]-x_2-2x_3=0[/MATH] but [MATH]x_3 = 0 \iff (1) = -x_2=1[/MATH] and [MATH](2)= -3x_2=1[/MATH] but this is clearly not true. Hence a = 0 is not a solution.
If a = 5, [MATH]x_2[/MATH]=t, then the solutions are given as: [MATH](x_1, x_2, x_3)[/MATH] = ([MATH]\frac{2}{5}(t-1)[/MATH], t, [MATH]1-2t)[/MATH]
If [MATH]a \neq 5[/MATH] then the solutions are given as: [MATH](x_1 ,x_2 ,x_3)[/MATH] = ([MATH]- \frac{2}{(a-3)a}[/MATH], [MATH]\frac{1}{a-3}[/MATH], 0)
Is this how I am supposed to answer the question / solve it or is there something lacking in my answer?