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(D) Three lines do not form a triangle if they are concurrent or if any two of them are parallel.Case-$1$: The lines are concurrent.The condition for

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$32$

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(D) Three lines do not form a triangle if they are concurrent or if any two of them are parallel.Case-$1$: The lines are concurrent.The condition for concurrency is that the determinant of the coefficients must be zero:$\begin{vmatrix} 2 & -1 & 3 \ 6 & 3 & 1 \ \alpha & 2 & -2 \end{vmatrix} = 0$$2(-6 - 2) - (-1)(-12 - \alpha) + 3(12 - 3\alpha) = 0$$2(-8) + 1(-12 - \alpha) + 3(12 - 3\alpha) = 0$$-16 - 12 - \alpha + 36 - 9\alpha = 0$$8 - 10\alpha = 0 \Rightarrow \alpha = \frac{4}{5}$.Case-$2$: Two lines are parallel.Line $L_1: 2x - y + 3 = 0$ (slope $m_1 = 2$)Line $L_2: 6x + 3y + 1 = 0$ (slope $m_2 = -2$)Line $L_3: \alpha x + 2y - 2 = 0$ (slope $m_3 = -\frac{\alpha}{2}$)$L_3$ is parallel to $L_1$ if $-\frac{\alpha}{2} = 2 \Rightarrow \alpha = -4$.$L_3$ is parallel to $L_2$ if $-\frac{\alpha}{2} = -2 \Rightarrow \alpha = 4$.The values of $\alpha$ are $\frac{4}{5}, 4, -4$.The sum of squares $p = (\frac{4}{5})^2 + (4)^2 + (-4)^2 = \frac{16}{25} + 16 + 16 = 32 + 0.64 = 32.64$.The greatest integer less than or equal to $p$ is $[32.64] = 32$.

If the sum of squares of all real values of $\alpha$,for which the lines $2x - y + 3 = 0$,$6x + 3y + 1 = 0$,and $\alpha x + 2y - 2 = 0$ do not form a triangle is $p$,then the greatest integer less than or equal to $p$ is $.........$

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