(c) Suppose that the position vectors of A, B, C, D, M and N are a, b, c, d, m and n respectively.
As, M and N are the mid-points of AC and BD.
So, m =
and n =
Then, AB + AD + CB + CD
= (b - a) + (d - a ) + (b - c)+ (d - c)
= 2(b + d) - 2(a + c)
= 2 x 2n - 2 x 2m = 4(n - m) = 4MN
⇒ 4λ + 1 - 7 - 2 - A = 10
⇒ 3 A = 18 ⇒ λ = 6