Animated Solution for Mathematics - Straight Lines: 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

Why Triangles Fail

Imagine standing on an infinite, flat plane. You have three lines, $L_1$, $L_2$, and $L_3$, stretching out in every direction.

In the vast majority of cases, these lines will dance across the plane, intersecting each other at three distinct points to form a beautiful, closed triangle. But today, we are looking for the rebels—the scenarios where these lines refuse to form a triangle. This is not just an algebraic exercise; it is a study of geometric collapse.

The Parallel Rebellion

Our first suspects are the parallel lines. If any two lines are parallel, they are destined to never meet, and without three intersection points, a triangle cannot exist.

We have our lines: $L_1:2x-y+3=0$ $L_2:6x+3y+1=0$ $L_3:\alpha x+2y-2=0$

Let us find their slopes. For $L_1$, we rewrite it as $y=2x+3$, giving us a slope $m_1=2$. For $L_2$, we rearrange $3y=-6x-1$ to get $y=-2x-\frac{1}{3}$, yielding $m_2=-2$.

Since $m_1 eq m_2$, $L_1$ and $L_2$ are not parallel. Now, we turn to $L_3$. Rearranging $\alpha x+2y-2=0$ gives $y=-\frac{\alpha }{2}x+1$, so $m_3=-\frac{\alpha }{2}$.

If $L_3$ is parallel to $L_1$, then $-\frac{\alpha }{2}=2$, which leads us to $\alpha =-4$. If $L_3$ is parallel to $L_2$, then $-\frac{\alpha }{2}=-2$, which gives us $\alpha =4$. These are our first two keys to the puzzle.

The Concurrency Trap

But what if the lines are not parallel? They could still fail to form a triangle if they all meet at the exact same point. This is called concurrency.

Imagine the three lines acting like spokes on a wheel, all crossing at a single hub. There is no space between them to enclose an area. To find this, we use the power of the determinant.

We set the determinant of the coefficients to zero:

Expanding this along the first row, we get:

Simplifying this, we have:

Combining like terms, we find $8-10\alpha =0$, or $10\alpha =8$, which simplifies to $\alpha =0.8$.

The Final Synthesis

We have found our three values for $\alpha$: $-4$, $4$, and $0.8$. The problem asks for the sum of the squares of these values, which we call $p$.

So, $p=(-4{)}^2+(4{)}^2+(0.8{)}^2=16+16+0.64=32.64$.

Finally, we seek the greatest integer less than or equal to $p$, denoted as $[p]$. Since $32.64$ sits comfortably between $32$ and $33$, the floor function gives us 32.

You have navigated the parallel paths and the concurrent traps to arrive at the truth. Geometry is not just about shapes; it is about understanding the conditions under which those shapes cease to be.