If the points with vectors $\alpha \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}, 6 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}, \frac{9}{2} \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$ are collinear, then $(19 \alpha-6 \beta)^2$ is equal to
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