= Solution
We use the <canonical height pairing> normalization $\langle P,Q\rangle=(\widehat h(P+Q)-\widehat h(P)-\widehat h(Q))/2$, so that $\langle P,P\rangle=\widehat h(P)$. The supplied values give
$$
\boxed{\langle P,Q\rangle\approx\frac{21.98-6.61-7.87}{2}=3.75.}
$$
The convention without the factor $1/2$ instead gives the doubled pairing $B(P,Q)\approx7.50$; all independence conclusions are the same.
In our convention the <Gram matrix> is approximately
$$
H=\begin{pmatrix}6.61&3.75\\3.75&7.87\end{pmatrix},\qquad
\det H\approx6.61\cdot7.87-3.75^2=37.9582>0.
$$
This positive margin is much larger than the rounding uncertainty. Even allowing each of the three stated heights an error of $0.01$, we have $a\geq6.60$, $b\geq7.86$ and $|c|\leq3.765$, giving $ab-c^2\geq6.60\cdot7.86-3.765^2=37.700775>0$. Thus the actual <canonical height pairing> matrix is <positive-definite>, not merely its rounded approximation.
If $rP+sQ=O$ for integers $r,s$, <bilinearity> and the <height parallelogram identity> give $0=\widehat h(rP+sQ)=ar^2+2crs+bs^2$. Positive definiteness forces $r=s=0$. This is the <rank certificate from rounded canonical heights>, and proves
$$
\boxed{\langle P,Q\rangle_{\mathrm{group}}\cong\mathbb Z\times\mathbb Z.}
$$
Here the last brackets denote the generated subgroup, not the scalar height pairing. The argument proves independence; it does not assert that these points generate the entire <Mordell-Weil group>.
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