Consider the system of linear equations
$-x+y+2 z=0$
$3 x-a y+5 z=1$
$2 x-2 y-a z=7$
Let $S_{1}$ be the set of all a $\in \mathbf{R}$ for which the system is inconsistent and $\mathrm{S}_{2}$ be the set of all $\mathrm{a} \in \mathbf{R}$ for which the system has infinitely many solutions. If $n\left(S_{1}\right)$ and $n\left(S_{2}\right)$ denote the number of elements in $S_{1}$ and $\mathrm{S}_{2}$ respectively, then
Correct Option: , 3
$\Delta=\left|\begin{array}{ccc}-1 & 1 & 2 \\ 3 & -\mathrm{a} & 5 \\ 2 & -2 & -\mathrm{a}\end{array}\right|$
$=-1\left(a^{2}+10\right)-1(-3 a-10)+2(-6+2 a)$
$=-a^{2}-10+3 a+10-12+4 a$
$\Delta=-a^{2}+7 a-12$
$\Delta=-\left[a^{2}-7 a+12\right]$
$\Delta=-[(a-3)(a-4)]$
$\Delta_{1}=\left|\begin{array}{ccc}0 & 1 & 2 \\ 1 & -\mathrm{a} & 5 \\ 7 & -2 & -\mathrm{a}\end{array}\right|$
$=0-1(-a-35)+2(-2+7 a)$
$\Rightarrow a+35-4+14 a$
$15 a+31$
Now
$\Delta_{1}=15 a+31$
For inconsistent $\Delta=0 \therefore \mathrm{a}=3, \mathrm{a}=4$ and for $\mathrm{a}=3$ and $4 \quad \Delta_{1} \neq 0$
$n\left(S_{1}\right)=2$
For infinite solution : $\Delta=0$ and $\Delta_{1}=\Delta_{2}=\Delta_{3}=0$
Not possible
$\therefore \mathrm{n}\left(\mathrm{S}_{2}\right)=0$
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