Let the position vectors of the points A, B, C and D be 5 ^ i + 5 ^...

Given that:

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

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

Now,

$\overset{\to}{AB} ≡ ( - 4, - 3, 3 - 2\lambda )$

$\overset{\to}{AC} ≡ ( - 7, \lambda - 5, 4 - 2\lambda )$

$\overset{\to}{AD} ≡ ( - 6, 0, 6 - 2\lambda )$

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

$∴ [ \overset{\to}{AB}\overset{\to}{AC}\overset{\to}{AD} ] = 0$

⇒ \left|\begin{cases}- 4 & - 3 & 3 - 2\lambda \\ - 7 & \lambda - 5 & 4 - 2\lambda \\ - 6 & 0 & 6 - 2\lambda\end{cases}\right| = 0

Expand the determinant along the first row.

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

On solving the above equation we get,

$∴ \lambda = 3, \lambda = 2$

We have two values of $\lambda$.

$⇒ \mathop{∑}_{\lambda \in s} ( \lambda + 2{)}^{2} = {\left(3 + 2\right)}^{2} + {\left(2 + 2\right)}^{2}$

$= 41$

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