Here is a second attempt to answer the question:

There are three equations for three unknown functions. As Igor points out, if you add the three equations together, you get $0 = 0$. Therefore, the third equation is a consequence of the first two. It therefore suffices to solve the first two equations only. This is now an underdetermined system, so there is a lot of freedom in what to do. Basically, you can set one function to anything you want and solve for the other two.

For certain values of the coefficients, the system could still be degenerate. If so, then there is another compatibility condition that must be satisfied by $b_1$ and $b_2$ and, if this condition is assumed to hold, the system reduces in a simiilar fashion to a single first order equation, which is essentially an ODE and therefore has a well-posed Cauchy problem.

If the system is nondegenerate, then you look for a constant linear transformation of the unknown functions $a_1, a_2, a_3$ into new unknown functions $u_1, u_2, u_3$ such that if you fix the function $u_3$ (i.e., set it to anything you want), the resulting $2$-by-$2$ system is hyperbolic. Recall that a $2$-by-$2$ first order differential operator in $3$ variables
$$
L_1\partial_1 + L_2\partial_2 + L_3\partial_3
$$
is hyperbolic if and only if the quadratic form $Q(x,y,z) = \det (xL_1 + yL_2 + zL_3)$ is nondegenerate and has indefinite signature, say $(-,+,+)$. Then the Cauchy problem is well-posed for an initial hypersurface if and only if $Q(v_1, v_2, v_3) > 0$ for any nonzero vector $(v_1,v_2,v_3)$ tangent to the hypersurface.

There are easier invariant ways to detect whether the system is hyperbolic or not, but I'll let someone else explain how to do that.

ADDED: Here is another way to describe what I said above.

1) If the left side of the three equations are scalar multiples of each other, then there is obviously only one PDE to solve and $b_1$, $b_2$ have to satisfy obvious consistency requirements.

2) Otherwise, two of the equations are not scalar multiples of each other but the third is a linear combination the other two. Let $E_1$ and $E_2$ be the two linearly independent equations and $E_3$ the third. To solve the whole system (i.e., all 3 equations), it suffices to solve $E_1$ and $E_2$ only. More generally, any two constants $p$ and $q$, it suffices to solve the equations $E_1 + pE_3$ and $E_2 + qE_3$. So the question reduces to whether you can find values of these constants so that this system is hyperbolic.

CORRECTION: If $p + q = 1$, then equations $E_1 + pE_3$ and $E_2 + qE_3$ are linearly dependent and therefore a solution to them is not necessarily a solution to the original system. Therefore, you must restrict to $p + q \ne 1$.

Below is my original incorrect answer:

The system is hyperbolic and has a well-posed Cauchy problem, if you can do the following: Find a linear change of co-ordinates $(z_1, z_2, z_3) = M(t, x, y)$, a linear change of basis for the unknown functions $(a_1, a_2, a_3) = L(u_1, u_2, u_3)$, and a linear change of basis for the eqautions $(b_1, b_2, b_3) = K(c_1, c_2, c_3)$ such that the system can be written in the form
$$
\partial_t u + A_1\partial_x u + A_2\partial_y u = c,
$$
such that one of the following holds (If so, then the Cauchy problem with initial data specified on the hypersurface $t = 0$ is well-posed):

- (strict hyperbolicity) There exists a matrix $Q(\xi,\eta)$ depending smoothly on $(\xi,\eta) \in \mathbb{R}^2$ such that the matrix $Q(\xi,\eta)(\xi A_1 + \eta A_2)Q^{-1}(\xi,\eta)$ is diagonal for all $(\xi,\eta)$.
- (symmetric hyperbolicity) The matrices $A_1$ and $A_2$ are symmetric.

A sufficient condition for strict hyperbolicity is the following: The matrix $\xi A_1 + \eta A_2$ has real, distinct eigenvalues for all $(\xi,\eta) \ne (0,0)$.