$=-1\left(a^{2}+10\right)-1(-3 a-10)+2(-6+2 a)$

$=-a^{2}-10+3 a+10-12+4 a$

$\Delta=-a^{2}+7 a-12$

$\Delta=-\left[a^{2}-7 a+12\right]$

$\Delta=-[(a-3)(a-4)]$

$\Delta_{1}=\left|\begin{array}{ccc}0 & 1 & 2 \\ 1 & -a & 5 \\ 7 & -2 & -a\end{array}\right|$

$=0-1(-\mathrm{a}-35)+2(-2+7 \mathrm{a})$

$\Rightarrow \mathrm{a}+35-4+14 \mathrm{a}$

$15 \mathrm{a}+31$

Now $\quad \Delta_{1}=15 \mathrm{a}+31$

For inconsistent $\Delta=0 \therefore \mathrm{a}=3, \mathrm{a}=4$

and for $\mathrm{a}=3$ and $4 \quad \Delta_{1} \neq 0$

$\mathrm{n}\left(\mathrm{S}_{1}\right)=2$

For infinite solution : $\Delta=0$

and $\Delta_{1}=\Delta_{2}=\Delta_{3}=0$

Not possible

$\therefore \mathrm{n}\left(\mathrm{S}_{2}\right)=0$