 Solution

The three handle attachments give $H_2(W;\mathbb{Z})\cong {\mathbb{Z}}^3$: there are no one-handles or three-handles. Use the capped core disks of handles as the homology basis, ordered by the framing labels $-2,-1,+1$. The surgery linking matrix gives the intersection form. Orient the three components so that the outer components have linking number $3$ and the middle component has linking number $1$ with each outer component. There are six crossings between the outer components, and two crossings between the middle component and each outer component; counting their signs gives these linking numbers. Reversing a component changes the corresponding row and column signs and does not change the invariants.

Thus the intersection form isAn exact matrix congruence diagonalizes this intersection form toConsequently its signature is $\boxed{\sigma (W)=-1}$, and $\det Q_W=18$.

The long exact sequence in relative homology and Poincare-Lefschetz duality identifywith the intersection form matrix, and identify $H_1(\partial W)$ with its cokernel, since $H_1(W)=0$. The Smith normal form has first invariant factor $1$, because the entries have greatest common divisor $1$, and second invariant factor $1$, because the upper-left two-by-two matrix minor has determinant $1$. The last invariant factor is $18$. Hence