The Levi-Civita symbol is defined as
$$
\varepsilon_{ijk} =
\begin{cases}
+1 & \text{if } (i,j,k) \text{ is } (1,2,3), (3,1,2) \text{ or } (2,3,1), \\
-1 & \text{if } (i,j,k) \text{ is } (1,3,2), (3,2,1) \text{ or } (2,1,3), \\
\;\;\,0 & \text{if }i=j \text{ or } j=k \text{ or } k=i
\end{cases}
$$
i.e. $\varepsilon_{ijk}$ is 1 if $(i, j, k)$ is an even permutations of $(1,2,3)$, $−1$ if it is an odd permutation, and $0$ if any index is repeated. For example
$$\begin{align}
\varepsilon_{132} &= -\varepsilon_{123} = - 1\\
\varepsilon_{312} &= -\varepsilon_{213} = -(-\varepsilon_{123}) = 1\\
\varepsilon_{231} &= -\varepsilon_{132} = -(-\varepsilon_{123}) = 1\\
\varepsilon_{232} &= -\varepsilon_{232} = 0
\end{align}
$$
Note that the denition implies that we are always free to cyclically permute indices
$$\varepsilon_{ijk} = \varepsilon_{kij} =\varepsilon_{jki}.$$
On the other hand, swapping any two indices gives a sign-change
$$\varepsilon_{ijk} = -\varepsilon_{ikj}.$$
The Kronecker delta is defined as:
$$
\delta_{ij} = \begin{cases}
0 &\text{if } i \neq j \\
1 &\text{if } i=j \end{cases}
$$
The Levi-Civita symbol is related to the Kronecker delta by the following equations
$$
\begin{align}
\varepsilon_{ijk}\varepsilon_{lmn} & = \begin{vmatrix} \delta_{il} & \delta_{im}& \delta_{in}\\ \delta_{jl} & \delta_{jm}& \delta_{jn}\\ \delta_{kl} & \delta_{km}& \delta_{kn}\\ \end{vmatrix}\\ & = \delta_{il}\left( \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}\right) - \delta_{im}\left( \delta_{jl}\delta_{kn} - \delta_{jn}\delta_{kl} \right) + \delta_{in} \left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl} \right).
\end{align}
$$
A special case of this result is (summing over $i$)
$$
\varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}.\tag 1
$$
The $i$th component of $\pmb a \times\pmb b$ is written as
$$
(\pmb a \times\pmb b)_i =
\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}a_jb_k = \varepsilon_{ijk}a_jb_k \tag 2
$$
where the last equality comes from the Einstein convention for repeated indices.
For example,
$$(\pmb a \times\pmb b)_1 = \varepsilon_{1jk}a_jb_k = \varepsilon_{123}a_2b_3 + \varepsilon_{132}a_3b_2 = a_2b_3 − a_3b_2.$$
Let us calculate the vector triple product $\pmb a \times (\pmb b \times \pmb c)$:
$$\begin{align}
(\pmb a \times (\pmb b \times \pmb c))_i
&= \varepsilon_{ijk}a_j(\pmb b \times \pmb c)_k \qquad\qquad\qquad\text{(use (2))}\\
&= (\varepsilon_{ijk}a_j)(\varepsilon_{klm}b_lc_m) \qquad\qquad\;\text{(use (2) with $ijk\to klm$)}\\
&= \varepsilon_{kij}\varepsilon_{klm}a_jb_lc_m\qquad\qquad\quad\;\;\;\text{(use $\varepsilon_{ijk} = \varepsilon_{kij}$)}\\
&=(\delta_{il}\delta_{jm} − \delta_{im}\delta_{jl})a_jb_lc_m\qquad\;\text{(use (1))}\\
&= a_jb_ic_j − a_jb_jc_i\qquad\qquad\quad\;\,\;\text{(use $\delta_{\mu\nu}\alpha_{\mu}\beta_{\nu}=\alpha_{\mu}\beta_{\mu}$)}\\
&= (\pmb a \cdot\pmb c)b_i − (\pmb a \cdot \pmb b)c_i\qquad\quad\;\;\,\;\text{(use $\alpha_{\mu}\beta_{\mu}=\pmb\alpha\cdot\pmb\beta$)}
\end{align}
$$
Therefore, we obtain
$$
\pmb a \times (\pmb b \times \pmb c)=(\pmb a \cdot \pmb c)\pmb b-(\pmb a \cdot \pmb b)\pmb c
$$
