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If the sum of squares of all real values of $\alpha$, for which the lines $2 x-y+3=0,6 x+3 y+1=0$ and $\alpha x+2 y-2=0$ do not form a triangle is $p$, then the greatest integer less than or equal to $\mathrm{p}$ is $.........$
$35$
$33$
$34$
$32$
Solution
$2 x-y+3=0 $
$ 6 x+3 y+1=0 $
$ \alpha x+2 y-2=0$
Will not form a $\Delta$ if $\alpha x+2 y-2=0$ is concurrent with $2 x-y+3=0$ and $6 x+3 y+1=0$ or parallel to either of them so
Case-$1$: Concurrent lines
$\left|\begin{array}{ccc}2 & -1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & -2\end{array}\right|=0 \Rightarrow \alpha=\frac{4}{5}$
Case-$2$ : Parallel lines
$ -\frac{\alpha}{2}=\frac{-6}{3} \text { or }-\frac{\alpha}{2}=2 $
$ \Rightarrow \alpha=4 \text { or } \alpha=-4 $
$ P=16+16+\frac{16}{25} $
$ {[\mathrm{P}]=\left[32+\frac{16}{25}\right]=32}$
