= Solution
Use a common intercept and laboratory-specific slopes in the <normal linear model>. With $t_i$ denoting days and $I_{Bi},I_{Ci}$ the <indicator variables> for laboratories B and C, the <design matrix> has row $x_i^T=(1,t_i,t_iI_{Bi},t_iI_{Ci})$, and
$$
Y_i=\beta_0+\beta_At_i+\delta_Bt_iI_{Bi}+\delta_Ct_iI_{Ci}+\varepsilon_i,
\qquad \varepsilon_i\overset{\mathrm{iid}}\sim N(0,\sigma^2).
$$
The <interaction terms> change slopes, without introducing laboratory-specific intercepts. Thus the estimated daily growth rates are
$$
\boxed{\widehat b_A=10.1404,\qquad\widehat b_B=10.1404+5.0770=15.2174,\qquad\widehat b_C=10.1404-3.1296=7.0108}
$$
in millimetres per day, with fitted common intercept $0.6464$ mm. The <residual degrees of freedom> are $30-4=26$. The usual unbiased error <variance> estimate is the square of the residual <standard error>:
$$
\boxed{s^2=15.09^2=227.7081\ \mathrm{mm}^2.}
$$
The small difference from $5921.7/26\approx227.758$ arises from rounding the <standard error> to two decimal places; the latter uses the more precise <residual sum of squares> supplied later. A <maximum-likelihood estimate> would instead divide that <residual sum of squares> by $30$.
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