The coordinates of the given points are $$A(2,5)$$ and $$B(6,1)$$.
Step 1: Equation of the perpendicular bisector $$L$$.
• Mid-point of $$AB$$: $$M\left(\dfrac{2+6}{2},\dfrac{5+1}{2}\right)=(4,3)$$.
• Slope of $$AB$$: $$m_{AB}=\dfrac{1-5}{6-2}=\dfrac{-4}{4}=-1$$.
• Hence slope of the perpendicular bisector is the negative reciprocal, i.e. $$+1$$.
Therefore $$L$$ is $$y-3 = 1(x-4) \;\Rightarrow\; y = x-1\;.$$
Step 2: Area condition for $$\triangle PAB$$.
Let $$P(x,y)$$ lie on $$L$$.
The area of a triangle formed by points $$P(x,y),\;A(2,5),\;B(6,1)$$ is
$$\text{Area}=\dfrac12
\left|\begin{vmatrix}
2 & 5 & 1\\
6 & 1 & 1\\
x & y & 1
\end{vmatrix}\right| \;.$$
Expanding the determinant,
$$\begin{aligned}
D &= 2(1-y) - 5(6-x) + 1(6y - x)\\
&= 2 - 2y - 30 + 5x + 6y - x\\
&= -28 + 4x + 4y.
\end{aligned}$$
The area is given to be $$10$$ square units, hence
$$\dfrac12|D| = 10 \;\Longrightarrow\; |-28 + 4x + 4y| = 20.$$
Dividing by 4,
$$|-7 + x + y| = 5 \;.\quad -(1)$$
Step 3: Impose the line condition $$y = x - 1$$.
Substitute $$y = x-1$$ into $$(1)$$:
$$\bigl| -7 + (x + x - 1) \bigr| = 5
\;\Longrightarrow\; |\,2x - 8\,| = 5.$$
Step 4: Solve for $$x$$ (two cases).
Case 1: $$2x - 8 =\; 5 \;\Rightarrow\; 2x = 13 \;\Rightarrow\; x = 6.5.$$
Then $$y = x - 1 = 5.5.$$
So $$P_1(6.5,\,5.5).$$
Case 2: $$2x - 8 = -5 \;\Rightarrow\; 2x = 3 \;\Rightarrow\; x = 1.5.$$
Then $$y = x - 1 = 0.5.$$
So $$P_2(1.5,\,0.5).$$
Step 5: Required sum of all the x and y coordinates.
Add the x- and y-coordinates of both admissible points:
$$\begin{aligned}
\text{Total} &= (6.5 + 5.5) + (1.5 + 0.5) \\
&= 12 + 2 = 14.
\end{aligned}$$
Hence the sum equals $$14$$.
Option C which is: $$14$$
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