= Solution
Three <vectors> in three-dimensional <Euclidean space> are non-coplanar precisely when their <scalar triple product> is nonzero. Put $\Delta=a\cdot(b\times c)\ne0$. In particular, $a\ne0$. The identity for <cross products with a common vector> gives
$$
(a\times b)\times(a\times c)=\bigl[a\cdot(b\times c)\bigr]a=\Delta a.
$$
For completeness, this follows from the <vector triple product identity>: writing $p=a\times b$ gives $p\times(a\times c)=a(p\cdot c)-c(p\cdot a)$, with $p\cdot c=\Delta$ and $p\cdot a=0$. Therefore the new <scalar triple product> is
$$
\boxed{a\cdot\bigl[(a\times b)\times(a\times c)\bigr]=|a|^2\Delta\ne0.}
$$
Thus $a,a\times b,a\times c$ are non-coplanar, or equivalently <linearly independent>.
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