I know how to solve the two-good case with $u(x) = \min\{x1, x2\}$, but the addition of $x3$ confuses me.

Problem

Derive the demand function $x(p,w)$ from $u(x) = \min\{x1, x2\} + x3$

What I did so far

We assume that in optimum $x1 = x2$.

Set up the budget constraint $p1x1 + p2x2 + p3x3 = w$

Rewrite budget constraint as $(p1+p2)x1 + p3x3 = w$ or $(p1+p2)x2 + p3x3 = w$

We can write $x1*$ or $x2* = \frac{w-p3x3}{p1+p2}$ and $x3* = \frac{w-(p1+p2)a}{p3}$

Confusion

How to proceed? Can I still use a Lagrangian to solve this?