= Solution
Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The <negative continued fraction> is
$$
\frac73=3-\frac1{2-1/2}=[3,2,2]^-.
$$
Attach three <two-handles> to a <zero-handle> along a chain of unknots, with adjacent <linking numbers> $+1$ and framings $-3,-2,-2$. Equivalently, take the <plumbing of oriented disk bundles> over three <spheres> with <Euler numbers> $-3,-2,-2$. Successive <slam-dunk moves> change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.
The zero sections are a basis of second <homology (mathematics)>, and the <intersection form> is
$$
\boxed{Q=\begin{pmatrix}-3&1&0\\1&-2&1\\0&1&-2\end{pmatrix}.}
$$
Its successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so <Sylvester's criterion> proves that $Q$ is negative definite. This explicitly constructs the required filling.
If <lens spaces> are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{smallmatrix}-2&1\\1&-4\end{smallmatrix}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.
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