If the circles ${x^2} + {y^2} + 6x + 8y + 16 = 0$ and ${x^2} + {y^2} + 2\left( {3 - \sqrt 3 } \right)x + 2\left( {4 - \sqrt 6 } \right)y = k + 6\sqrt 3 + 8\sqrt 6$, $k > 0$, touch internally at the point $P(\alpha ,\beta )$, then ${\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2}$ is equal to ________________.