A pole has to be erected on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. The distance of the pole from one of the gates is:

Let us construct a diagram based on the given statements.
There are 2 diametrically opposite gates A and B.
A pole is erected on the circumference such that the distance of the pole from one of the gates is 7 m more than the distance of the pole from the other gate. Let the distances be x and x+7 m.

Now, APB is a right-angled triangle (since AB is the diameter).
Applying Pythagoras theorem, we get,
$$13^2 = x^2 + (x+7)^2$$
$$169=x^2+x^2+14x+49$$
$$120 = 2x^2 + 14x$$
$$x^2 + 7x - 60 = 0$$
$$(x+12)(x-5)=0$$
Therefore, $$x$$ can be $$-12$$ or $$5$$.
$$x$$ cannot be negative. Therefore, $$x$$ has to be $$5$$.
The distance of the pole from one of the gates is 5 m. Therefore, option C is the right answer.