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Step-by-Step Solution
Step 1: Write down the condition for the area of the triangle
Let the points be $P(5,\,6)$, $Q(3,\,2)$, and $R(\alpha,\,\beta)$. The area of triangle $PQR$ is given by
$$
\text{Area} \;=\; \frac{1}{2}\,\bigl|\,x_1(y_2 - y_3) \;+\; x_2(y_3 - y_1) \;+\; x_3(y_1 - y_2)\bigr|.
$$
Substituting $P(5,6)$, $Q(3,2)$, and $R(\alpha,\beta)$ into this formula, we want the area to be $12$ square units.
Step 2: Set up the absolute value equation
Using the formula for the area of a triangle in coordinate geometry,
$$
12 \;=\; \frac{1}{2}\,\Bigl|\,5(2 - \beta) \;+\; 3(\beta - 6) \;+\; \alpha\,(6-2)\Bigr|.
$$
Simplify inside the absolute value:
$$
5(2-\beta) \;=\; 10 \;-\; 5\beta, \quad
3(\beta-6) \;=\; 3\beta \;-\; 18, \quad
\alpha(6-2) \;=\; 4\alpha.
$$
Summing these:
$$
10 - 5\beta + 3\beta - 18 + 4\alpha \;=\; -8 - 2\beta + 4\alpha.
$$
Hence,
$$
12 \;=\; \frac{1}{2}\,\bigl|\,4\alpha - 2\beta - 8\bigr|.
$$
Therefore,
$$
24 \;=\; \bigl|\,4\alpha - 2\beta - 8\bigr|.
$$
Step 3: Solve for the lines describing set A
The above absolute value yields two linear equations:
\[
4\alpha - 2\beta - 8 = 24 \quad \text{or} \quad 4\alpha - 2\beta - 8 = -24.
\]
Simplifying each:
1) $4\alpha - 2\beta = 32 \quad\Longrightarrow\quad 2\alpha - \beta = 16.$
2) $4\alpha - 2\beta = -16 \quad\Longrightarrow\quad 2\alpha - \beta = -8.$
Thus, the set $A$ of all points $(\alpha,\beta)$ satisfying the area condition is given by the union of the lines
$$
2\alpha - \beta = 16 \quad\text{and}\quad 2\alpha - \beta = -8.
$$
Step 4: Compute the distance from the origin to each line
We want the least possible length of a line segment from the origin $(0,0)$ to any point on these lines.
The distance $d$ from a point $(x_0,y_0)$ to a line $ax + by + c = 0$ is
$$
d = \frac{\bigl|a x_0 + b y_0 + c\bigr|}{\sqrt{a^2 + b^2}}.
$$
• For $2\alpha - \beta = 16$: rewrite as $2x - y - 16 = 0$. Distance from the origin:
$$
d_1 = \frac{\lvert0 - 0 - 16\rvert}{\sqrt{2^2 + (-1)^2}} \;=\; \frac{16}{\sqrt{5}}.
$$
• For $2\alpha - \beta = -8$: rewrite as $2x - y + 8 = 0$. Distance from the origin:
$$
d_2 = \frac{\lvert0 - 0 + 8\rvert}{\sqrt{2^2 + (-1)^2}} \;=\; \frac{8}{\sqrt{5}}.
$$
Step 5: Identify the minimum distance
Among the two distances, $d_2 = \frac{8}{\sqrt{5}}$ is the smaller one. Hence, the least possible length of a segment joining the origin to a point in $A$ is
$$
\frac{8}{\sqrt{5}}.
$$
This matches the given correct answer.
Final Answer
$\displaystyle \frac{8}{\sqrt{5}}$
