Key Concepts and Formulas

• Condition for Three Lines Not Forming a Triangle: Three lines do not form a triangle if they are concurrent or if at least two of them are parallel.
• Condition for Concurrency: Three lines $A_1x + B_1y + C_1 = 0$, $A_2x + B_2y + C_2 = 0$, and $A_3x + B_3y + C_3 = 0$ are concurrent if $\begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0$.
• Condition for Parallel Lines: Two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are parallel if their slopes are equal, i.e., $-\frac{A_1}{B_1} = -\frac{A_2}{B_2}$.

Step-by-Step Solution

Step 1: Define the lines and check for initial parallelism

We are given the three lines: Line 1 ($L_1$): $2x - y + 3 = 0$ Line 2 ($L_2$): $6x + 3y + 1 = 0$ Line 3 ($L_3$): $\alpha x + 2y - 2 = 0$

First, let's find the slopes of $L_1$ and $L_2$ to check if they are parallel. Slope of $L_1$, $m_1 = -\frac{2}{-1} = 2$. Slope of $L_2$, $m_2 = -\frac{6}{3} = -2$.

Since $m_1 \neq m_2$, lines $L_1$ and $L_2$ are not parallel. Thus, we need to consider concurrency and the parallelism of $L_3$ with either $L_1$ or $L_2$.

Step 2: Case 1 - Concurrency

For the three lines to be concurrent, the determinant of their coefficients must be zero: $\begin{vmatrix} 2 & -1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & -2 \end{vmatrix} = 0$

Expanding the determinant: $2(3(-2) - 1(2)) - (-1)(6(-2) - 1(\alpha)) + 3(6(2) - 3(\alpha)) = 0$ $2(-6 - 2) + 1(-12 - \alpha) + 3(12 - 3\alpha) = 0$ $2(-8) - 12 - \alpha + 36 - 9\alpha = 0$ $-16 - 12 - \alpha + 36 - 9\alpha = 0$ $8 - 10\alpha = 0$ $10\alpha = 8$ $\alpha = \frac{8}{10} = \frac{4}{5}$ So, one value of $\alpha$ is $\alpha_1 = \frac{4}{5}$.

Step 3: Case 2 - $L_3$ parallel to $L_1$

The slope of $L_3$ is $m_3 = -\frac{\alpha}{2}$. For $L_3$ to be parallel to $L_1$, we must have $m_3 = m_1$, so: $-\frac{\alpha}{2} = 2$ $\alpha = -4$ So, another value of $\alpha$ is $\alpha_2 = -4$.

Step 4: Case 3 - $L_3$ parallel to $L_2$

For $L_3$ to be parallel to $L_2$, we must have $m_3 = m_2$, so: $-\frac{\alpha}{2} = -2$ $\alpha = 4$ So, a third value of $\alpha$ is $\alpha_3 = 4$.

Step 5: Calculate the sum of squares

The values of $\alpha$ are $\frac{4}{5}$, $-4$, and $4$. The sum of their squares is: $p = \left(\frac{4}{5}\right)^2 + (-4)^2 + (4)^2 = \frac{16}{25} + 16 + 16 = \frac{16}{25} + 32 = \frac{16 + 32 \cdot 25}{25} = \frac{16 + 800}{25} = \frac{816}{25} = 32.64$

Step 6: Find the greatest integer less than or equal to $p$

We need to find $\lfloor p \rfloor = \lfloor 32.64 \rfloor = 32$.

Common Mistakes & Tips

• Carefully expand the determinant, paying attention to signs.
• Remember to check for parallelism between all possible pairs of lines.
• Double-check arithmetic calculations to avoid errors.

Summary

We considered the conditions for three lines not forming a triangle: concurrency and parallelism. We found three values of $\alpha$ that satisfy these conditions: $\frac{4}{5}$, $-4$, and $4$. The sum of the squares of these values is $32.64$, and the greatest integer less than or equal to this sum is $32$.

The final answer is $\boxed{32}$.