Let $$k$$ be an integer such that the triangle with vertices $$(k, -3k)$$, $$(5, k)$$ and $$(-k, 2)$$ has area 28 sq. units. Then the orthocenter of this triangle is at the point:

Let us denote the three given vertices by

$$A(k,\,-3k),\qquad B(5,\,k),\qquad C(-k,\,2).$$

The area of a triangle with vertices $$(x_1,y_1),\,(x_2,y_2),\,(x_3,y_3)$$ is obtained from the determinant formula

$$\text{Area}=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$

Substituting $$(x_1,y_1)=(k,-3k),\;(x_2,y_2)=(5,k),\;(x_3,y_3)=(-k,2)$$ we have

$$\begin{aligned} x_1(y_2-y_3)&=k\,(k-2)=k^2-2k,\\[2mm] x_2(y_3-y_1)&=5\bigl(2-(-3k)\bigr)=5(2+3k)=10+15k,\\[2mm] x_3(y_1-y_2)&=-k\bigl((-3k)-k\bigr)=-k(-4k)=4k^2. \end{aligned}$$

Adding these three results gives

$$k^2-2k+10+15k+4k^2=5k^2+13k+10.$$

Because the area is given as 28 square units, we must have

$$\dfrac12\bigl|5k^2+13k+10\bigr|=28\;\;\Longrightarrow\;\;\bigl|5k^2+13k+10\bigr|=56.$$

This absolute-value equation splits into two quadratic equations.

1. $$5k^2+13k+10=56\;\Longrightarrow\;5k^2+13k-46=0.$$

2. $$5k^2+13k+10=-56\;\Longrightarrow\;5k^2+13k+66=0.$$

The second quadratic has discriminant

$$\Delta_2=13^2-4\cdot5\cdot66=169-1320=-1151<0,$$

so it possesses no real roots and can be ignored. We therefore solve the first quadratic.

For $$5k^2+13k-46=0,$$ the discriminant is

$$\Delta_1=13^2-4\cdot5\cdot(-46)=169+920=1089=33^2.$$

Hence, using the quadratic formula

$$k=\dfrac{-13\pm33}{2\cdot5}=\dfrac{-13\pm33}{10}.$$

The two possible values are

$$k=\dfrac{20}{10}=2,\qquad k=\dfrac{-46}{10}=-\dfrac{23}{5}.$$

Because $$k$$ is required to be an integer, only $$k=2$$ is admissible.

Thus the triangle’s concrete vertices are

$$A(2,-6),\qquad B(5,2),\qquad C(-2,2).$$

To locate the orthocenter we intersect two of the altitudes of the triangle.

Altitude from vertex A:

The side $$BC$$ runs from $$B(5,2)$$ to $$C(-2,2).$$ Its slope is

$$m_{BC}=\dfrac{2-2}{5-(-2)}=0,$$

so $$BC$$ is the horizontal line $$y=2.$$ The altitude from $$A$$ is therefore perpendicular to a horizontal line, meaning it is vertical through $$x=x_A=2.$$ Hence the first altitude has equation

$$x=2.$$

Altitude from vertex B:

The side $$AC$$ joins $$A(2,-6)$$ and $$C(-2,2).$$ Its slope is

$$m_{AC}=\dfrac{2-(-6)}{-2-2}=\dfrac{8}{-4}=-2.$$

An altitude is perpendicular to the side, so the required slope is the negative reciprocal

$$m_{\perp B}=\dfrac12.$$

Passing through $$B(5,2)$$, the altitude from $$B$$ is written as

$$y-2=\dfrac12\,(x-5).$$

This rearranges to

$$y=\dfrac12x-\dfrac12.$$

Intersection of the two altitudes:

We simultaneously satisfy

$$x=2\qquad\text{and}\qquad y=\dfrac12x-\dfrac12.$$

Substituting $$x=2$$ into the second equation,

$$y=\dfrac12\cdot2-\dfrac12=1-\dfrac12=\dfrac12.$$

Hence the orthocenter is

$$H\bigl(2,\;\dfrac12\bigr).$$

The coordinates $$\left(2,\;\dfrac12\right)$$ correspond to Option D.

Hence, the correct answer is Option D.