1. Let the vertices be

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

We are given that the area of triangle $ABC$ is $28$.

2. Use the coordinate area formula

Area of triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is

$\Delta=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$

Substituting:

$\Delta=\frac12\left|k(k-2)+5(2+3k)+(-k)(-3k-k)\right|.$

Now simplify:

$k(k-2)=k^2-2k,$ $5(2+3k)=10+15k,$ $(-k)(-4k)=4k^2.$

So,

$\Delta=\frac12\left|k^2-2k+10+15k+4k^2\right|$ $=\frac12|5k^2+13k+10|.$

Given area $=28$,

$\frac12|5k^2+13k+10|=28$ $|5k^2+13k+10|=56.$

Hence,

$5k^2+13k+10=56 \quad \text{or} \quad 5k^2+13k+10=-56.$

That gives:

$5k^2+13k-46=0$ or $5k^2+13k+66=0.$

The second has negative discriminant, so no real solution.

For the first:

$5k^2+13k-46=0.$

Discriminant:

$D=13^2-4\cdot 5\cdot(-46)=169+920=1089=33^2.$

Thus,

$k=\frac{-13\pm 33}{10}.$

So,

$k=2 \quad \text{or} \quad k=-\frac{23}{5}.$

Since $k$ is an integer,

$\boxed{k=2}.$

3. Find the triangle vertices for $k=2$

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

4. Find two altitudes

Altitude from $A$

Since $B(5,2)$ and $C(-2,2)$ have same $y$-coordinate, line $BC$ is

$y=2,$ which is horizontal.

So the altitude from $A$ is vertical:

$x=2.$

Altitude from $B$

Slope of $AC$ is

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

So slope of altitude from $B$ is the negative reciprocal:

$m=\frac12.$

Equation through $B(5,2)$:

$y-2=\frac12(x-5).$

5. Find the orthocentre

Orthocentre is intersection of the altitudes.

From altitude from $A$:

$x=2.$

Substitute into altitude from $B$:

$y-2=\frac12(2-5)=\frac12(-3)=-\frac32.$

So,

$y=2-\frac32=\frac12.$

Hence the orthocentre is

$\boxed{\left(2,\frac12\right)}.$

6. Match with the options

This is Option C.

7. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They agree.