Let α β γ=45 ; α, β, γ ∈ ℝ. If x(α, 1,2)+y(1, β, 2)+z(2,3, γ)=(0,0,0) for some x, y, z ∈ ℝ, x y z ≠ 0, then 6 α+4 β+γ is equal to _________.

Solution

Maths sir · 4 steps

Three vectors combine to zero with nonzero weights, so they are coplanar and their determinant is zero.

1. Read the condition. The vectors (α,1,2), (1,β,2) and (2,3,γ) combine to the zero vector with x, y, z not all zero. So they are linearly dependent, which means they are coplanar. Their determinant must therefore be zero.

   |α, 1, 2; 1, β, 2; 2, 3, γ| = 0

2. Expand the determinant. Expand along the first row: α(βγ - 6) - 1(γ - 4) + 2(3 - 2β) = 0. This simplifies to αβγ - 6α - γ + 4 + 6 - 4β = 0.
3. Use the product. Put in αβγ = 45 to get 45 + 10 - (6α + 4β + γ) = 0.
4. So. 6α + 4β + γ = 55, so the answer is 55.

The idea to keep. If three vectors add to zero with nonzero coefficients, they are linearly dependent, so they are coplanar. Set their determinant to zero and the required relation usually drops straight out.

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