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
\(\left|\begin{array}{lll}\alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma\end{array}\right|=0\)
\( \Rightarrow \alpha(\beta \gamma-6)-1(\gamma-6)+2(2-2 \beta)=0\)
\(\Rightarrow \alpha \beta \gamma-6 \alpha-\gamma+6+4-4 \beta=0\)
\(\Rightarrow 6 \alpha+4 \beta+\gamma=55\)