The element $u$ must have weight $-\alpha_2$, since $\mu = \lambda - \alpha_2.$ 

In $U(\mathfrak{n^-})$ there are only two linearly independent elements that have such weight (assuming PBW basis with respect to fixed order of generators based on positive roots): $y_{\alpha_2}$ and $y_{\alpha_1}y_{\alpha_3}.$ Hence the sought element $u$ is a linear combination of such
$$
u = a y_{\alpha_2}  + b y_{\alpha_1}y_{\alpha_3}.
$$

Since this has to be the image of the higest weight vector of $M(\mu)$ we must have $x_{\alpha_1} u = 0$ and $x_{\alpha_3} u = 0.$ Writing it out and using relations defining the Verma module and the Lie algebra, we end up with system of 2 linear equations for 2 unknowns. E.g. we have
$$
x_{\alpha_1} (ay_{\alpha_2}v_\lambda) = (a[x_{\alpha_1}, y_{\alpha_2}] + ay_{\alpha_2} x_{\alpha_1})v_\lambda
$$
where the first term on the right hand side is either zero, or some element of Cartan subalgebra acting on $v_\lambda$, and the second term is zero from the definition of the Verma module.