I have found three different ways to write the extrinsic curvature tensor,
$\begin{align*} K_{ab}^{(1)} &= \frac{1}{2}P^c_a P^d_b\mathcal{L}_n g_{cd}, \\ K_{ab}^{(2)} &= P^c_aP^d_b \nabla_{(c} n_{c)}, \\ K_{ab}^{(3)} &= \nabla_a n_b - n_a n^\lambda \nabla_\lambda n_b. \end{align*}$
where $P^a_b$ is the projection tensor and $n^a$ is the unit normal vector.
I understand how to go from $K_{ab}^{(1)}$ to $K_{ab}^{(2)}$ and these are clearly symmetric tensors. However, what I'm struggling with is how to show that $K_{ab}^{(3)}$ is a symmetric tensor.
I have found a hint in a book that says to use the fact that $n^a$ is hypersurface orthogonal and then $$\begin{align*} n^a K_{ab} &= n^a \nabla_a n_b - n^a n_a n^\lambda \nabla_\lambda n_b \\ &= A_a - A_a =0 \end{align*}$$ where $A_a$ is the acceleration defined as $A^a = n^b \nabla_b n^a,$ but I'm failing to see how this shows that $K_{ab}^{(3)}$ is symmetric.
How can I see that $K_{ab}^{(3)}$ is a symmetric tensor?
First observe also that $K_{ab}^{(3)} n^b \equiv 0$ since the derivative of any constant is zero. So it suffices to look at the tangential components.
Let $X, Y$ be two vector fields tangent to your hypersurface. Compute
$$ X^a Y^b K_{ab}^{(3)} - X^b Y^a K_{ab}^{(3)} = g(\nabla_X n, Y) - g(\nabla_Y n, X)$$
Since $g(n,Y) = g(n,X) = 0$ you have $g(\nabla_X n, Y) = - g(n, \nabla_X Y)$ etc. This implies
$$ X^a Y^b K_{ab}^{(3)} - X^b Y^a K_{ab}^{(3)} = g(n, [Y,X]) $$
since the Levi-Civita connection is torsion free. Since $X,Y$ are both tangent to a hypersurface, their commutator $[Y,X]$ is also tangent to the same hypersurface (quick proof: let $f$ be a defining function of your hypersurface. You know that $X(f) = Y(f) = 0$, and hence $[Y,X](f) = 0$). This implies the final term on the right vanishes. Hence $K^{(3)}$ is symmetric.