If the area of the triangle whose one vertex is at the vertex of the parabola, $$y^2 + 4(x - a^2) = 0$$ and the other two vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is:

We have the parabola $$y^{2}+4(x-a^{2})=0.$$

First we rewrite it in the standard horizontal‐opening form. Moving the term $$4(x-a^{2})$$ to the other side gives

$$y^{2}=-4(x-a^{2}).$$

So we may write

$$y^{2}=-4x+4a^{2}\quad\Longrightarrow\quad x=a^{2}-\frac{y^{2}}{4}.$$

Comparing with the standard equation $$y^{2}=4p(x-h)$$, we see that the vertex is at $$\bigl(a^{2},\,0\bigr).$$

Now we find the points where the parabola meets the y-axis. On the y-axis we have $$x=0,$$ so we substitute $$x=0$$ in the original equation:

$$y^{2}+4(0-a^{2})=0\quad\Longrightarrow\quad y^{2}-4a^{2}=0.$$

This factors as

$$y^{2}=4a^{2}\quad\Longrightarrow\quad y=\pm2a.$$

Thus the two intersection points are $$\bigl(0,\,2a\bigr)$$ and $$\bigl(0,\,-2a\bigr).$$

Hence the three vertices of the required triangle are

$$V\bigl(a^{2},0\bigr),\;P_{1}\bigl(0,2a\bigr),\;P_{2}\bigl(0,-2a\bigr).$$

To obtain the area, we use the shoelace (determinant) formula for a triangle whose vertices are $$\bigl(x_{1},y_{1}\bigr),\bigl(x_{2},y_{2}\bigr),\bigl(x_{3},y_{3}\bigr):$$

$$\text{Area}=\frac12\,\Bigl|\,x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\,\Bigr|.$$

Assigning the coordinates, we take

$$\bigl(x_{1},y_{1}\bigr)=\bigl(a^{2},0\bigr),\quad \bigl(x_{2},y_{2}\bigr)=\bigl(0,2a\bigr),\quad \bigl(x_{3},y_{3}\bigr)=\bigl(0,-2a\bigr).$$

Now we compute each term:

$$x_{1}(y_{2}-y_{3})=a^{2}\bigl(2a-(-2a)\bigr)=a^{2}\cdot4a=4a^{3},$$

$$x_{2}(y_{3}-y_{1})=0\cdot\bigl(-2a-0\bigr)=0,$$

$$x_{3}(y_{1}-y_{2})=0\cdot\bigl(0-2a\bigr)=0.$$

Adding these gives

$$x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})=4a^{3}.$$

Therefore the area is

$$\text{Area}=\frac12\,\bigl|\,4a^{3}\bigr|=2\,|a^{3}|.$$

We are told that this area equals $$250$$ square units, so

$$2\,|a^{3}|=250\quad\Longrightarrow\quad|a^{3}|=125.$$

Taking the real cube root (remembering that $$|a^{3}|=(|a|)^{3}$$), we get

$$|a|=125^{1/3}=5.$$

Thus the parameter $$a$$ can be $$5$$ or $$-5.$$ Among the given options, only the positive value appears, so we select

$$a=5.$$

Hence, the correct answer is Option D.