(55)

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

$\Rightarrow \alpha x+y+2z=0; x+\beta y+3z=0; 2x+2y+\gamma z=0$

Since, $xyz\ne 0$ so the system of equations has non-trivial solution.

Now, $\left|\begin{array}{ccc}\alpha & 1& 2\\ 1& \beta & 3\\ 2& 2& \gamma \end{array}\right|$$=0$

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

$\Rightarrow \alpha \beta \gamma -6\alpha -\gamma +6+4-4\beta =0$

$\Rightarrow 45+10-6\alpha -\gamma -4\beta =0\Rightarrow 6\alpha +4\beta +\gamma =55$

(38)

Given, $2x+7y+\lambda z=3$

             $3x+2y+5z=4$

             $x+\mu y+32z=-1$ has infinitely many solutions.

$\Rightarrow \Delta =0,{\Delta }_1=0,{\Delta }_2=0$ and ${\Delta }_3=0$

${\Delta }_2=$$\left|\begin{array}{ccc}2& \lambda & 3\\ 3& 5& 4\\ 1& 32& -1\end{array}\right|$$=0$ and ${\Delta }_3=$$\left|\begin{array}{ccc}2& 7& 3\\ 3& 2& 4\\ 1& \mu & -1\end{array}\right|$$=0$

$\Rightarrow -266+\lambda \left(7\right)+273=0$ and $-4-8\mu +49+9\mu -6=0$

$\Rightarrow 7\lambda +7=0$ and $\mu +39=0$

$\Rightarrow \lambda =-1$ and $\mu =-39$

So, $\lambda -\mu =-1+39=38$