Let the position vectors of the points A,B,C,D be 5i^ +5j^+2λk^, i^+2j^+3k^, −2i^+λj^+4k^ and −i^+5j^+6k^ Let the set S ={λ∈ R : The points A, B,C and D are coplanar} then ∑λ∈s (λ+2)2 is equal to

[AB¯AC¯AD¯]=0⇒λ2-5λ+6=0⇒λ=2,3∑λ∈s(λ+2)2=(2+2)2+(3+2)2=41

Let the position vectors of the points A,B,C,D be 5i^ +5j^+2λk^, i^+2j^+3k^, −2i^+λj^+4k^ and −i^+5j^+6k^ Let the set S ={λ∈ R : The points A, B,C and D are coplanar} then ∑λ∈s (λ+2)2 is equal to

[AB¯AC¯AD¯]=0⇒λ2-5λ+6=0⇒λ=2,3∑λ∈s(λ+2)2=(2+2)2+(3+2)2=41

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Let the position vectors of the points A,B,C,D be $5 \hat{i} + 5 \hat{j} + 2 λ \hat{k}, \hat{i} + 2 \hat{j} + 3 \hat{k}, - 2 \hat{i} + λ \hat{j} + 4 \hat{k} a n d - \hat{i} + 5 \hat{j} + 6 \hat{k}$
Let the set S ={ $λ \in R :$ The points A, B,C and D are coplanar} then $\sum_{λ \in s} {(λ + 2)}^{2}$ is equal to

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$\frac{37}{2}$

answer is B.

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$[\bar{A B} \bar{A C} \bar{A D}]$ $= 0 \Rightarrow λ^{2} - 5 λ + 6 = 0 \Rightarrow λ = 2, 3$

$\sum_{λ \in s} (λ + 2)^{2} = (2 + 2)^{2} + (3 + 2)^{2} = 41$

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Let the position vectors of the points A,B,C,D be 5i^ +5j^+2λk^, i^+2j^+3k^, −2i^+λj^+4k^ and −i^+5j^+6k^ Let the set S ={λ∈ R : The points A, B,C and D are coplanar} then ∑λ∈s (λ+2)2 is equal to

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