= Solution
For the unheaded continuation, write the <vector triple product> as
$$
[\mathbf a\times(\mathbf b\times\mathbf c)]_i=\epsilon_{ijk}a_j\epsilon_{klm}b_lc_m.
$$
The <Levi-Civita symbol> contraction is $\epsilon_{ijk}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$. One way to check its sign is to fix distinct $i,j$: the only nonzero term has $k$ the remaining index, and the pairs $(l,m)=(i,j),(j,i)$ contribute $+1,-1$ respectively. When $i=j$, both sides vanish. Contracting with the <vectors> now gives both requested notations:
$$
\boxed{[\mathbf a\times(\mathbf b\times\mathbf c)]_i=b_i a_jc_j-c_i a_jb_j,\qquad \mathbf a\times(\mathbf b\times\mathbf c)=\mathbf b(\mathbf a\cdot\mathbf c)-\mathbf c(\mathbf a\cdot\mathbf b).}
$$
If $\mathbf a$ and $\mathbf b$ are <orthogonal>, then $\mathbf a\cdot\mathbf b=0$. With the specified $\mathbf c$, the perpendicularity of the <cross product> gives $\mathbf a\cdot\mathbf c=|\mathbf a|^2$. Therefore
$$
\boxed{\mathbf a\times(\mathbf b\times\mathbf c)=|\mathbf a|^2\mathbf b.}
$$
The same answer covers the degenerate case in which one of these two <vectors> is zero.
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