= Solution
Three <vectors> form a <basis> of three-dimensional space exactly when their <scalar triple product> is nonzero:
$$
\boxed{\Delta=\mathbf u\cdot(\mathbf v\times\mathbf w)\ne0}.
$$
It is the <determinant> of the <matrix> with these columns, so this condition is equivalent to <linear independence> and spanning the whole space.
Take the <dot product> of the expansion of $\mathbf y$ with $\mathbf v\times\mathbf w$. The terms containing $\mathbf v$ and $\mathbf w$ vanish, leaving $\mathbf y\cdot(\mathbf v\times\mathbf w)=\alpha\Delta$. Cycling the same argument yields
$$
\boxed{\alpha=\frac{\mathbf y\cdot(\mathbf v\times\mathbf w)}{\Delta},\qquad
\beta=\frac{\mathbf y\cdot(\mathbf w\times\mathbf u)}{\Delta},\qquad
\gamma=\frac{\mathbf y\cdot(\mathbf u\times\mathbf v)}{\Delta}}.
$$
The three cross-product numerators use the <reciprocal basis>.
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