Linear Expenditure System of Demands, Derivation Help
I am not sure I understand your question. $\lambda$ is a Lagrange multiplicator which has a value in the optimum. It is not a parameter and hence you cannot leave it in your solution.
 When solving the Lagrangian the optimal solution has the form $(x,\lambda)$ and in addition to your conditions
$$x_1 = \frac{\alpha}{\lambda p_1} + b_1$$
$$x_2 = \frac{\beta}{\lambda p_2} + b_2$$
$$x_3 = \frac{\gamma}{\lambda p_3} + b_3$$
it also has to fulfill
$$
(p \cdot x - w) \lambda = 0.
$$
If $\lambda \neq 0$ this also means $p \cdot x = w$. $\lambda$ cannot be equal to zero because it is in the denominator of your optimal solution for $x$:
$$x(p,w) = (b_1, b_2, b_3) + \left(\frac{\alpha}{\lambda p_1},\frac{\beta}{\lambda p_2},\frac{\gamma}{\lambda p_3}\right)$$
and hence you were correct to use the Walras law or budget constraint.

For more on optimality conditions see the Karush-Kuhn-Tucker theorem.

About $b_1, b_2$ and $b_3$: I think this is supposed to be a bliss point.