How to compute the Marshallian demand for this specific utility function $$u(x) = x_1^\alpha \cdot (x_2+x_3)^{1-\alpha}, \text{ with } \alpha \in(0,1)$$

I tried to set up the Lagrangian and it turns out $\lambda$ have no solution unless $p_1 = p_2$.

$$L = x_1^\alpha \cdot (x_2+x_3)^{1-\alpha} -\lambda(x_1 p_1 + x_2 p_2 + x_3 p_3 -1)$$

$$\frac{\partial L}{\partial x_2} = (1-\alpha)x_1^\alpha(x_2+x_3)^{-\alpha} -\lambda p_2 = 0$$

$$\frac{\partial L}{\partial x_3} = (1-\alpha)x_1^\alpha(x_2+x_3)^{-\alpha} -\lambda p_3 = 0$$

I noticed this function is similar to C-D utility function. So I'm wondering if I can combine $x_{3}$ and $x_{2}$ as one good and then apply C-S utility function. But this method seems unreliable to me. 

Any one can help?
It means that the level surfaces of $u$ don't touch the surface $x_1p_1+x_2p_2+x_3p_3=1$ unless $p_2=p_3$. But, since $x_i\ge0$, it follows that the admissible domain of $x_i$ is compact and the maximum is still achieved somewhere. Namely, in some boundary point of the  admissible domain $-$ for $x_2=0$ or $x_3=0$.