Let the position vectors of the points A,B,C and D be 5 i+5 j+2λ k, i+2 j+3 k,-2 i+λ j+4 k and - i+5 j+6 k. Let the set S=λ ∈ mathbbR: the points…

Given that:

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

$D(-1,5,6)$D(−1,5,6).

Now,

$$\vec{AB}\equiv (-4,-3,3-2\lambda )$$AB≡(−4,−3,3−2λ)

$$\vec{AC}\equiv (-7,\lambda -5,4-2\lambda )$$AC≡(−7,λ−5,4−2λ)

$$\vec{AD}\equiv (-6,0,6-2\lambda )$$AD≡(−6,0,6−2λ)

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

$$\therefore [\vec{AB}\vec{AC}\vec{AD}]=0$$∴[ABACAD]=0

$$\Rightarrow |\begin{matrix}-4 & -3 & 3-2\lambda \\ -7 & \lambda -5 & 4-2\lambda \\ -6 & 0 & 6-2\lambda \end{matrix}|=0$$⇒∣−4−7−6​−3λ−50​3−2λ4−2λ6−2λ​∣=0

Expand the determinant along the first row.

$$\Rightarrow -4[(\lambda -5)(6-2\lambda )-0]-(-3)[-7(6-2\lambda )-(-6)(4-2\lambda )]+(3-2\lambda )[0-(-6)(\lambda -5)]$$⇒−4[(λ−5)(6−2λ)−0]−(−3)[−7(6−2λ)−(−6)(4−2λ)]+(3−2λ)[0−(−6)(λ−5)]

On solving the above equation we get,

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

We have two values of $\lambda$λ.

$$\Rightarrow \underset{\lambda \in s}{\sum }(\lambda +2{)}^{2}={(3+2)}^{2}+{(2+2)}^{2}$$⇒λ∈s∑​(λ+2)2=(3+2)2+(2+2)2

$$=41$$=41

Hence, $\displaystyle \underset{\lambda \in s}{\sum }(\lambda +2{)}^{2}=41$λ∈s∑​(λ+2)2=41