Let the position vectors of the points A , B , C and D be 5 i ^ + 5 j ^ + 2 λ k ^ , i ^ + 2 j ^ + 3 k ^…

Given that:

$A\mathit{(}5,5,2\lambda \mathit{)}$, $B\left(1,2,3\right)$, $C\left(-2,\lambda ,4\right)$ and 

$D\left(-1,5,6\right)$.

Now,

$\overrightarrow{AB}\equiv (-4,-3,3-2\lambda )$

$\overrightarrow{AC}\equiv (-7,\lambda -5,4-2\lambda )$

$\overrightarrow{AD}\equiv (-6,0,6-2\lambda )$

Given that $A,B,C \text{and} D$ are coplanar.

$\therefore [\overrightarrow{AB} \overrightarrow{AC} \overrightarrow{AD}]=0$

$\Rightarrow \left|\begin{array}{ccc}-4& -3& 3-2\lambda \\ -7& \lambda -5& 4-2\lambda \\ -6& 0& 6-2\lambda \end{array}\right|=0$

Expand the determinant along the first row.

$\Rightarrow -4\left[\left(\lambda -5\right)\left(6-2\lambda \right)-0\right]-\left(-3\right)\left[-7\left(6-2\lambda \right)-\left(-6\right)\left(4-2\lambda \right)\right]+\left(3-2\lambda \right)\left[0-\left(-6\right)\left(\lambda -5\right)\right]$

On solving the above equation we get,

$\therefore \lambda =3,\lambda =2$

We have two values of $\lambda$.

$\Rightarrow \mathit{\sum }_{\lambda \mathit{\in }s}(\lambda +2{)}^2={\left(3+2\right)}^2+{\left(2+2\right)}^2$

$=41$

Hence,  $\mathit{\sum }_{\lambda \mathit{\in }s}(\lambda +2{)}^2=41$