Let the position vectors of the points A, B, C and D be  $5\widehat{\mathrm{i}}+5\widehat{\mathrm{j}}+2\mathrm{\lambda }\widehat{\mathrm{k}},\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+3\widehat{\mathrm{k}},-2\widehat{\mathrm{i}}+\mathrm{\lambda }\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}$ and $-\widehat{\mathrm{i}}+5\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}$. Let the set $\mathrm{S}=\{\mathrm{\lambda }\in \mathrm{R}$: the points A, B, C and D are coplanar}. Then $\sum _{\mathrm{\lambda }\in \mathrm{S}} (\mathrm{\lambda }+2{)}^2$is equal to

Given points are $\mathrm{A}(5,3,2\mathrm{\lambda })\mathrm{B}(1,2,3);\mathrm{C}(-2,\mathrm{\lambda },4);\mathrm{D}(-1,5,6)$
$\overrightarrow{AB}=(-4,-3,3-2\lambda )$
$$\begin{gathered} \begin{array}{l}\overrightarrow{AC}=(-7,\lambda -5,4-2\lambda ) \\ \overrightarrow{AD}=(-6,0,6-2\lambda )\end{array} \end{gathered}$$
$\begin{array}{l}The points A, B, C, D are co-planar\\ \therefore [\overrightarrow{AB} \overrightarrow{AC} \overrightarrow{AD}]=0\\ \left|\begin{array}{ccc}-4& -3& 3-2\lambda \\ -7& \lambda -5& 4-2\lambda \\ -6& 0& 6-2\lambda \end{array}\right|=0\\ \Rightarrow -4(\lambda -3)(\lambda -2)=0\\ \therefore \lambda =3,\lambda =2\\ \sum _{\lambda \in S} (\lambda +2{)}^2=5^2+4^2\\ =41\end{array}$