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

$\begin{aligned} & \alpha \beta \gamma=45, \alpha \beta \gamma \in \mathrm{R} \\ & \mathrm{x}(\alpha, 1,2)+\mathrm{y}(1, \beta, 2)+\mathrm{z}(2,3, \gamma)=(0,0,0) \\ & \mathrm{x}, \mathrm{y}, \mathrm{z} \in \mathrm{R}, \mathrm{xyz} \neq 0 \\ & \alpha \mathrm{x}+\mathrm{y}+2 \mathrm{z}=0 \\ & \mathrm{x}+\beta \mathrm{y}+3 \mathrm{z}=0 \\ & 2 \mathrm{x}+2 \mathrm{y}+\gamma \mathrm{z}=0 \\ & \mathrm{xyz} \neq 0 \Rightarrow \text { non-trivial } \\ & \left|\begin{array}{lll}\alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma\end{array}\right|=0\end{aligned}$​αβγ=45,αβγ∈Rx(α,1,2)+y(1,β,2)+z(2,3,γ)=(0,0,0)x,y,z∈R,xyz=0αx+y+2z=0x+βy+3z=02x+2y+γz=0xyz=0⇒ non-trivial ​α12​1β2​23γ​​=0​ $\begin{aligned} & \Rightarrow \alpha(\beta \gamma-6)-1(\gamma-6)+2(2-2 \beta)=0 \\ & \Rightarrow \alpha \beta \gamma-6 \alpha-\gamma+6+4-4 \beta=0 \\ & \Rightarrow 6 \alpha+4 \beta+\gamma=55\end{aligned}$​⇒α(βγ−6)−1(γ−6)+2(2−2β)=0⇒αβγ−6α−γ+6+4−4β=0⇒6α+4β+γ=55​