The scalar product of the vector a = i + j + k with a unit vector along the sum of vectors b = 2i + 4j - k and c = λi + 2j + 3k is equal to one. Find the value of λ and hence find the unit vector along b + c.

Given:
a = i + j + k
b = 2i + 4j - k
c = λi + 2j + 3k
So, b + c = (2 + λ)i + 6j + 2k
Unit vector along b + c = [(2 + λ)i + 6j + 2k] / √[(2 + λ)² + 36 + 4] = [(2 + λ)i + 6j + 2k] / √[(2 + λ)² + 40]
Given that dot product of a with the unit vector of b + c is equal to 1.
So, apply given condition,
(2 + λ) + 6 + 2 / √[(2 + λ)² + 40] = 1
=> 8 + λ = √[(2 + λ)² + 40]
Squaring, we get
64 + λ² + 16λ = 4 + λ² + 4λ + 40
=> 12λ = -20
=> λ = -5/3