= Solution
Let $\widehat\delta_D$ and $\widehat\delta_I$ be <independent>, approximately unbiased estimates of one consistent treatment effect, with <variances> $v_D,v_I>0$. The <inverse-variance weighted mean> gives
$$
\boxed{\widehat\delta=\frac{\widehat\delta_D/v_D+\widehat\delta_I/v_I}{1/v_D+1/v_I},\qquad \operatorname{Var}(\widehat\delta)=\left(\frac1{v_D}+\frac1{v_I}\right)^{-1}.}
$$
Indeed, weights $w,1-w$ give <variance> $w^2v_D+(1-w)^2v_I$; differentiating gives $w=v_I/(v_D+v_I)$, the weight on the direct estimate. Here the three displayed trials give $\widehat\delta_D=-1.33$, $v_D=0.04$, and hence $\widehat\delta\simeq-1.3203$ with <variance> $0.03448$. This numerical pooling uses the displayed evidence, not the unavailable data from all twelve trials. A shared study or shared control would introduce <covariance>, requiring the corresponding correlated-estimator formula; inconsistency would invalidate the common-effect interpretation.
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