The two given equations represent planes, and the required line is their intersection. They can be written in vector form as
$$(x,y,z)\cdot U = 8$$
$$(x,y,z)\cdot V = 15$$
where $U = (1,1,-1)$ and $V = (2,2,1)$ are vectors that are normal to the two planes.
It is obvious (I think) that the line is parallel to the cross product vector $U \times V$.
So, we can use $(a_2, b_2, c_2) = U \times V$.
Next, we need some point $(a_1,b_1,c_1)$ on the line. There are many possible choices, of course. In effect, we just need to choose some third plane and intersect our line with this. In other words, we add a third linear equation to the two we already have, and solve. One easy choice is the plane $x=0$. Using this equation together with the original two gives the solution $(a_1,b_1,c_1)=(0,23/3,-1/3)$.