Let →a, →b, and →c be three non-coplanar unit vectors such that the angle between every pair of them is π/3. If →a × →b + →b × →c = p→a + q→b + r→c, where p, q, and r are scalars, then the value of p² + 2q² + r²/q² is?

|→a| = |→b| = |→c| = 1
→a × →b + →b × →c = p→a + q→b + r→c
Taking dot product with →a on both sides:
→a ⋅ (→a × →b + →b × →c) = →a ⋅ (p→a + q→b + r→c)
0 + →a ⋅ (→b × →c) = p|→a|² + q(→a ⋅ →b) + r(→a ⋅ →c)
Since |→a| = |→b| = |→c| = 1 and the angle between every pair is π/3, we have:
→a ⋅ →b = →a ⋅ →c = →b ⋅ →c = cos(π/3) = 1/2
Also, →a ⋅ (→b × →c) represents the scalar triple product [→a →b →c], which is given by:
[→a →b →c] = →a ⋅ (→b × →c) = |→a| |→b| |→c| sin(θ) where θ is the angle between →b × →c and →a.
The volume of the parallelepiped formed by →a, →b, →c is given by V = |[→a →b →c]| = |→a ⋅ (→b × →c)|
Using the formula for the scalar triple product for a unit cube:
[→a →b →c] = |→a||→b||→c|(1 - cos²(π/3))^(1/2) = 1 * 1 * 1 * (1 - (1/2)²)^(1/2) = √3/2
Therefore, √3/2 = p + q(1/2) + r(1/2)
Taking dot product with →b on both sides:
→b ⋅ (→a × →b + →b × →c) = →b ⋅ (p→a + q→b + r→c)
→b ⋅ (→a × →b) + →b ⋅ (→b × →c) = p(→b ⋅ →a) + q(→b ⋅ →b) + r(→b ⋅ →c)
0 + 0 = p(1/2) + q + r(1/2)
Taking dot product with →c on both sides:
→c ⋅ (→a × →b + →b × →c) = →c ⋅ (p→a + q→b + r→c)
→c ⋅ (→a × →b) + →c ⋅ (→b × →c) = p(→c ⋅ →a) + q(→c ⋅ →b) + r(→c ⋅ →c)
√3/2 + 0 = p(1/2) + q(1/2) + r
We have the following system of equations:
p + q/2 + r/2 = √3/2
p/2 + q + r/2 = 0
p/2 + q/2 + r = √3/2
Solving these equations, we get p = 0, q = -√3/2, r = √3/2
Therefore, p² + 2q² + r²/q² = 0 + 2(3/4) + (3/4)/(3/4) = 3/2 + 1 = 5/2
However, there seems to be a mistake in the calculation or the question itself. Let's re-examine.