Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in \mathbb{R}$. If $x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3, \gamma)=(0,0,0)$ for some $x, y, z \in \mathbb{R}, x y z
eq 0$, then $6 \alpha+4 \beta+\gamma$ is equal to _______

Question

Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in \mathbb{R}$. If $x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3, \gamma)=(0,0,0)$ for some $x, y, z \in \mathbb{R}, x y z 
eq 0$, then $6 \alpha+4 \beta+\gamma$ is equal to _______

Aicharya · Accepted Answer

$\begin{aligned} & \alpha \beta \gamma=45, \alpha \beta \gamma \in \mathrm{R} \ & \mathrm{x}(\alpha, 1,2)+\mathrm{y}(1, \beta, 2)+\mathrm{z}(2,3, \gamma)=(0,0,0) \ & \mathrm{x}, \mathrm{y}, \mathrm{z} \in \mathrm{R}, \mathrm{xyz} 
eq 0 \ & \alpha \mathrm{x}+\mathrm{y}+2 \mathrm{z}=0 \ & \mathrm{x}+\beta \mathrm{y}+3 \mathrm{z}=0 \ & 2 \mathrm{x}+2 \mathrm{y}+\gamma \mathrm{z}=0 \ & \mathrm{xyz} 
eq 0 \Rightarrow 	ext { non-trivial } \ & \left|\begin{array}{lll}\alpha & 1 & 2 \ 1 & \beta & 3 \ 2 & 2 & \gamma\end{array}ight|=0\end{aligned}$ $\begin{aligned} & \Rightarrow \alpha(\beta \gamma-6)-1(\gamma-6)+2(2-2 \beta)=0 \ & \Rightarrow \alpha \beta \gamma-6 \alpha-\gamma+6+4-4 \beta=0 \ & \Rightarrow 6 \alpha+4 \beta+\gamma=55\end{aligned}$

Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in \mathbb{R}$. If $x(\alpha, 1,2)+y(1, \beta, 2)+z(2,3…

Solution