If (1, 5, 35), (7, 5, 5), (1, λ, 7) and (2λ, 1, 2) are coplanar, then the sum of all possible values of λ is :

Solution

Maths sir · 4 steps

Fix one point, draw three vectors to the other points, and set their box product to zero.

1. Build three vectors. Take A(1, 5, 35) as the base point. Then AB⃗ = (6, 0, -30), AC⃗ = (0, λ - 5, -28) and AD⃗ = (2λ - 1, -4, -33). The four points are coplanar exactly when these three vectors lie in one plane, that is, when [AB⃗ AC⃗ AD⃗] = 0.

   |6, 0, -30; 0, λ - 5, -28; 2λ - 1, -4, -33| = 0

2. Expand. Take 6 out of the first row, which leaves (1, 0, -5). Expanding along that row gives [-33(λ - 5) - 112] - 5[-(λ - 5)(2λ - 1)] = 0. This simplifies to 10λ² - 88λ + 78 = 0, or 5λ² - 44λ + 39 = 0.
3. Find the roots. The quadratic factorises as (5λ - 39)(λ - 1) = 0, so λ = 1 or λ = 39/5. Both roots are real, so both values are allowed.
4. So. The sum of the possible values is 1 + 39/5 = 44/5, which is option (c).

The idea to keep. For four coplanar points, fix one and form three vectors to the others. Their scalar triple product must be 0. For a sum of values, use -b/a once you know the roots are real.

Answer from the NTA final answer key. Solution by PrepWiser.