Question

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 $p$, then the greatest integer less than or equal to $p$ is ....

Answer 1

Given:
$2x - y + 3 = 0, \quad 6x + 3y + 1 = 0, \quad ax + 2y - 2 = 0.$

To not form a triangle, $ax + 2y - 2 = 0$ must be concurrent or parallel with the other lines.

Solving for concurrent lines:
$\frac{2}{6} = \frac{-1}{3} \implies a = \frac{4}{5}.$

Similarly, for parallel lines:
$a = \pm 4.$

Calculating $p$:
$p = \left(\frac{4}{5}\right)^2 + 4^2 + 4^2 = 32.$

The Correct answer is: 32