If (2,3,9),(5,2,1),(1, λ, 8) and (λ, 2,3) are coplanar, then the product of all possible values of λ is:

Solution

Maths sir · 4 steps

Fix one point, draw three vectors from it, and set their scalar triple product to zero.

1. Three vectors. Take A(2,3,9) as the base point. Then AB⃗ = (3,-1,-8), AC⃗ = (-1,λ-3,-1) and AD⃗ = (λ-2,-1,-6). The four points are coplanar exactly when these three vectors lie in one plane.

   |3, -1, -8; -1, λ-3, -1; λ-2, -1, -6| = 0

2. Expand the determinant. Expanding along the first row gives 3(-6λ+17) + (λ+4) - 8(-λ²+5λ-5). That is -18λ+51+λ+4+8λ²-40λ+40.
3. The quadratic. Collecting terms gives 8λ² - 57λ + 95 = 0. You do not need the roots themselves. The product of the roots is c/a = 95/8.
4. So. The product of all possible values of λ is 95/8, which is option (d).

The idea to keep. Four points are coplanar when the triple product of the three vectors from one of them is zero. If you get a quadratic in λ, read the sum or product straight from its coefficients.

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