If the sum of squares of all real values of α , for which the lines...

Given lines are $2x - y + 3 = 0$, $6x + 3y + 1 = 0$ and $\alpha x + 2y - 2 = 0$.

These lines will not form a triangle if $\alpha x + 2y - 2 = 0$ is concurrent with $2x - y + 3 = 0$ and $6x + 3y + 1 = 0$ or parallel to either of them.

Case-$1$: Concurrent lines

⇒ \left|\begin{cases}2 & - 1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & - 2\end{cases}\right| = 0

$⇒ 2\left(- 6 - 2\right) + 1\left(- 12 - \alpha\right) + 3\left(12 - 3\alpha\right) = 0$

$⇒ - 16 - 12 - \alpha + 36 - 9\alpha = 0$

$⇒ - 10\alpha + 8 = 0$

$⇒ \alpha = \dfrac{4}{5}$

Case-$2$: Parallel lines

$\dfrac{- \alpha}{6} = \dfrac{- 2}{3}$, $- \dfrac{\alpha}{2} = \dfrac{- 2}{- 1}$

$⇒ \alpha = 4$, $\alpha = - 4$

$⇒ P = 16 + 16 + \dfrac{16}{25}$

$[ P ] = \left[32 + \dfrac{16}{25}\right] = 32$