Let the position vectors of the points A,B,C and D be 5iˆ+5jˆ+2λkˆ,iˆ+2jˆ+3kˆ,−2iˆ+λjˆ+4kˆ and −iˆ+5jˆ+6kˆ. Let the set S={λ∈R: the points A,B,C and D are coplanar}. The ∑λ∈S(λ+2)2 is equal to

The given position vectors are:

A = 5i + 5j + 2λk B = i + 2j + 3k C = -2i + λj + 4k D = -i + 5j + 6k

For the points A, B, C, and D to be coplanar, the scalar triple product of the vectors AB, AC, and AD must be zero.

The vector AB = B - A = -4i - 3j + (3-2λ)k The vector AC = C - A = -7i + (λ-5)j + (4-2λ)k The vector AD = D - A = -6i + j + (6-2λ)k

The scalar triple product of AB, AC, and AD is given by the determinant of the matrix with AB, AC, and AD as its rows:

| -4 -3 3-2λ | | -7 λ-5 4-2λ | | -6 0 6-2λ |

This determinant must be equal to zero.

Calculating the determinant, we get:

-4[(λ-5)(6-2λ) - 0] - (-3)[-7(6-2λ) - (-6)(4-2λ)] + (3-2λ)[-7*0 - (-6)(λ-5)]

Solving this equation will give us the values of λ for which the points A, B, C, and D are coplanar.

Once we have the values of λ, we can substitute them into the expression (λ+2)^2 and sum the results to get the final answer.

The given points A, B, C, and D are coplanar if the volume of the parallelepiped formed by them is zero. The volume V of the parallelepiped formed by the vectors AB, AC, and AD is given by the scalar triple product:

V = AB . (AC x AD)

where "." denotes the dot product and "x" denotes the cross product.

First, we need to find the vectors AB, AC, and AD:

AB = B - A = iˆ+2jˆ+3kˆ - (5iˆ+5jˆ+2λkˆ) = -4iˆ - 3jˆ + (3-2λ)kˆ AC = C - A = -2iˆ+λjˆ+4kˆ - (5iˆ+5jˆ+2λkˆ) = -7iˆ + (λ-5)jˆ + (4-2λ)kˆ AD = D - A = -iˆ+5jˆ+6kˆ - (5iˆ+5jˆ+2λkˆ) = -6iˆ + jˆ + (6-2λ)kˆ

Next, we find the cross product AC x AD:

AC x AD = (-7iˆ + (λ-5)jˆ + (4-2λ)kˆ) x (-6iˆ + jˆ + (6-2λ)kˆ) = ((λ-5)(6-2λ) - (4-2λ))iˆ - ((-7)(6-2λ) - (-6)(4-2λ))jˆ + ((-7) - (-6)(λ-5))kˆ = (6λ-30-4λ+10)iˆ + (42λ-168+24λ-48)jˆ + (-7+6λ+30)kˆ = (2λ-20)iˆ + (66λ-216)jˆ + (6λ+23)kˆ

Finally, we find the dot product AB . (AC x AD):

AB . (AC x AD) = -4*(2λ-20) - 3*(66λ-216) + (3-2λ)*(6λ+23) = -8λ + 80 - 198λ + 648 + 18λ - 46λ - 69 = -234λ + 659

Setting this equal to zero gives the values of λ for which the points A, B, C, and D are coplanar:

-234λ + 659 = 0 λ = 659 / 234 ≈ 2.816

Therefore, the set S = {2.816}. The sum ∑λ∈S(λ+2)2 is then (2.816+2)^2 = 4.816^2 = 23.197.