The centre of family of circles cutting the family of circles $x^2+y^2+4x\left(\lambda -\frac{3}{2}\right)+3y\left(\lambda -\frac{4}{3}\right)-6\left(\lambda +2\right)=0$ orthogonally, lies on

$A$ circle is inscribed (i.e., touches all four sides ) into a rhombous $ABCD$ with one angle $60°.$The distance from the centre of the circle to the nearest vertex is equal to $1.$ If $P$ is any point of the circle, then $|PA{|}^2+|PB{|}^2+|PC{|}^2+|PD{|}^2$ is equal to