Let αβγ = 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

Question

Let \(\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in R. If \,x(\alpha, 1,2)+y(1, \beta, 2)\) \( +\mathrm{z}(2,3, \gamma)=(0,0,0)\) for some x, y, z ∈ R, xyz ≠ 0, then 6α + 4β + γ is equal to ---------

Answer 1

Correct answer is : 55

\(\alpha \beta \gamma=45, \alpha \beta \gamma \in R\)

\(x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3, \gamma)=(0,0,0)\)

\(\mathrm{x}, \mathrm{y}, \mathrm{z} \in \mathrm{R}, \mathrm{xyz} \neq 0\)

\(\alpha x+y+2 z=0\)

\(x+\beta y+3 z=0\)

\(2 \mathrm{x}+2 \mathrm{y}+\gamma \mathrm{z}=0\)

\(\mathrm{xyz} \neq 0 \Rightarrow \)non-trivial

\(\left|\begin{array}{lll}\alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma\end{array}\right|=0\)

\( \Rightarrow \alpha(\beta \gamma-6)-1(\gamma-6)+2(2-2 \beta)=0\)

\(\Rightarrow \alpha \beta \gamma-6 \alpha-\gamma+6+4-4 \beta=0\)

\(\Rightarrow 6 \alpha+4 \beta+\gamma=55\)