homework and exercises - Derivation of Christoffel Symbols

So I am reading a book on relativity & differential geometry and in the text, they gave the Christoffel symbols in terms of the metric and its derivatives, but I wanted to derive it myself. However, when I derived it, I seem to be missing two terms. Can somebody spot where I messed up?

From the text, they said that the derivative of the basis vectors $\vec{e}_{\mu}$, denoted as $\vec{e}_{\mu, \nu} \equiv \partial_{\nu}\vec{e}_{\mu}$, can be written as a linear combination of these basis vectors and also a normal vector, i.e. $$\vec{e}_{\mu,\nu}=\Gamma_{\mu\nu}^{\lambda}\vec{e}_{\lambda}+K_{\mu \nu} \vec{n}$$

I also know that the metric itself, $g_{\mu \nu}$ can be written as the dot product of these basic vectors as $$g_{\mu \nu} = \vec{e}_{\mu} \cdot \vec{e}_{\nu}$$

So my logic was to take the derivative of the metric with this definition:

$$\begin{split}\partial_{\alpha} g_{\mu \nu} & =\partial_{\alpha} (\vec{e}_{\mu} \cdot \vec{e}_{\nu}) \\ & =\partial_{\alpha}\vec{e}_{\mu} \cdot \vec{e}_{\nu} + \vec{e}_{\mu} \cdot \partial_{\alpha} \vec{e}_{\nu} \\ & =\vec{e}_{\mu,\alpha} \cdot \vec{e}_{\nu} + \vec{e}_{\mu} \cdot \vec{e}_{\nu, \alpha} \\ & =(\Gamma_{\mu \alpha}^{\lambda} \vec{e}_{\lambda} +K_{\mu \alpha} \vec{n} ) \cdot \vec{e}_{\nu} +\vec{e}_{\mu} \cdot (\Gamma_{\nu \alpha}^{\lambda} \vec{e}_{\lambda} + K_{\nu \alpha} \vec{n}) \\ & =\Gamma_{\mu \alpha}^{\lambda} (\vec{e}_{\lambda} \cdot \vec{e}_{\nu}) + \Gamma_{\nu \alpha}^{\lambda} (\vec{e}_{\mu} \cdot \vec{e}_{\lambda}) \\ & =\Gamma_{\mu \alpha}^{\lambda} g_{\lambda \nu} + \Gamma_{\nu \alpha}^{\lambda} g_{\mu \lambda} \\ & =\Gamma_{\mu \alpha}^{\lambda} g_{\lambda \nu} + \Gamma_{\nu \alpha}^{\lambda} g_{\lambda \mu} \end{split}$$

In this, the only thing I used was that $\vec{n} \cdot \vec{e}_{\lambda} =0$ by definition and that the metric is symmetric, i.e. $g_{\mu \lambda} = g_{\lambda \mu}$.

So now that I have that equation for the derivative of the metric, I might as well play around with it and solve for the Christoffel symbols. The only thing I did was multiply the whole equation by $g^{\alpha \lambda}$ in an attempt to contract and eliminate some of the metric terms to isolate $\Gamma$:

$$\begin{split} g^{\alpha \lambda} \partial_{\alpha} g_{\mu \nu} & = \Gamma_{\mu \alpha}^{\lambda} g^{\alpha \lambda} g_{\lambda \nu} + \Gamma_{\nu \alpha}^{\lambda} g^{\alpha \lambda} g_{\lambda \nu} \\ & =\Gamma_{\mu \alpha}^{\lambda} \delta_{\nu}^{\alpha} + \Gamma_{\nu \alpha}^{\lambda} \delta_{\mu}^{\alpha} \end{split}$$

Since this is just multiplying the metric by its inverse, it results in the identity matrix, or the Kronecker delta. Since this is $0$ when the indices are not equal to each other and $1$ when they are, we can write this as:

$$g^{\alpha \lambda} \partial_{\alpha} g_{\mu \nu} = \Gamma_{\mu \nu}^{\lambda} + \Gamma_{\nu \mu}^{\lambda}$$

And lastly the Christoffel symbols are symmetric in their lower two indices so we finally get:

$$g^{\alpha \lambda} \partial_{\alpha} g_{\mu \nu} = 2 \Gamma_{\mu \nu}^{\lambda}$$ or $$\Gamma_{\mu \nu}^{\lambda} = \frac{1}{2} g^{\alpha \lambda} (\partial_{\alpha} g_{\mu \nu})$$

The problem is that the actual (correct) answer for $\Gamma$ involves three derivatives of the metric instead of my one. Where have I gone wrong here?