1. For three lines to not form a triangle, at least one of the following must happen:

   • some two lines are parallel, or
   • all three lines are concurrent.

   Let the lines be: $L_1: 2x-y+3=0$ $L_2: 6x+3y+1=0$ $L_3: \alpha x+2y-2=0$

2. First, check whether $L_1$ and $L_2$ are parallel.

   Write in slope form: $L_1: y=2x+3 \quad \Rightarrow \text{slope}=2$ $L_2: 3y=-6x-1 \Rightarrow y=-2x-\frac13 \quad \Rightarrow \text{slope}=-2$

   So $L_1$ and $L_2$ are not parallel.

3. Find values of $\alpha$ such that $L_3$ is parallel to $L_1$.

   For $L_3$: $2y=-\alpha x+2 \Rightarrow y=-\frac{\alpha}{2}x+1$ So slope of $L_3$ is $-\frac{\alpha}{2}$.

   Parallel to $L_1$ means: $-\frac{\alpha}{2}=2$ $\alpha=-4$

4. Find values of $\alpha$ such that $L_3$ is parallel to $L_2$.

   Parallel to $L_2$ means: $-\frac{\alpha}{2}=-2$ $\alpha=4$

5. Now check concurrency of all three lines.

   First find intersection point of $L_1$ and $L_2$.

   From $L_1$: $y=2x+3$

   Substitute into $L_2$: $6x+3(2x+3)+1=0$ $6x+6x+9+1=0$ $12x+10=0$ $x=-\frac56$

   Then $y=2\left(-\frac56\right)+3=-\frac53+3=\frac43$

   So intersection point is $\left(-\frac56,\frac43\right)$

6. For concurrency, this point must satisfy $L_3$: $\alpha x+2y-2=0$ $\alpha\left(-\frac56\right)+2\left(\frac43\right)-2=0$ $-\frac{5\alpha}{6}+\frac83-2=0$ $-\frac{5\alpha}{6}+\frac23=0$ $-5\alpha+4=0$ $\alpha=\frac45$

7. Therefore all real values of $\alpha$ for which the three lines do not form a triangle are: $\alpha=-4,\ 4,\ \frac45$

8. Sum of squares of these values: $p=(-4)^2+4^2+\left(\frac45\right)^2$ $p=16+16+\frac{16}{25}$ $p=32+\frac{16}{25}=\frac{816}{25}=32.64$

9. Hence, $\lfloor p \rfloor = \lfloor 32.64 \rfloor = 32$

So the required integer is $32$.