$$ABC$$ is a triangle, right angled at $$A$$. The resultant of the forces acting along $$\overrightarrow{AB}, \overrightarrow{AC}$$ with magnitudes $$\dfrac{1}{AB}$$ and $$\dfrac{1}{AC}$$ respectively is the force along $$\overrightarrow{AD}$$, where $$D$$ is the foot of the perpendicular from $$A$$ onto $$BC$$. The magnitude of the resultant is

Place the triangle in a rectangular coordinate system with $$A$$ at the origin $$O(0,0)$$.
Choose the positive $$x$$-axis along $$\overrightarrow{AB}$$ and the positive $$y$$-axis along $$\overrightarrow{AC}$$.

With this choice we have
$$B(AB,0),\qquad C(0,AC).$$

The two given forces are

$$\mathbf{F}_1=\dfrac{1}{AB}\,\hat{i},\qquad \mathbf{F}_2=\dfrac{1}{AC}\,\hat{j},$$
because the unit vectors along $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$ are $$\hat{i}$$ and $$\hat{j}$$, respectively.

Their resultant is

$$\mathbf{R}=\mathbf{F}_1+\mathbf{F}_2=\left(\dfrac{1}{AB},\,\dfrac{1}{AC}\right).$$

Next find the direction of $$\overrightarrow{AD}$$, where $$D$$ is the foot of the perpendicular from $$A$$ onto $$BC$$.
The equation of the hypotenuse $$BC$$ through $$B(AB,0)$$ and $$C(0,AC)$$ is obtained by two-point form:

$$\dfrac{x}{AB}+\dfrac{y}{AC}=1\qquad -(1)$$

Equation $$-(1)$$ has the form $$px+qy=1$$ with
$$p=\dfrac{1}{AB},\quad q=\dfrac{1}{AC}.$$

The vector normal to line $$-(1)$$ is $$\mathbf{n}=(p,q)=\left(\dfrac{1}{AB},\,\dfrac{1}{AC}\right).$$
A perpendicular dropped from the origin to the line travels in the direction of this normal, so $$\overrightarrow{AD}$$ is collinear with $$\mathbf{n}$$.

But $$\mathbf{n}$$ is exactly the resultant vector $$\mathbf{R}$$ we found earlier. Hence the resultant force acts along $$\overrightarrow{AD}$$, as stated in the question.

The magnitude of the resultant is therefore

$$\lvert\mathbf{R}\rvert=\sqrt{\left(\dfrac{1}{AB}\right)^2+\left(\dfrac{1}{AC}\right)^2}.$$

The perpendicular distance from the origin to line $$-(1)$$ (which is the length $$AD$$) is given by the standard distance-from-point-to-line formula:

$$AD=\frac{\lvert1\rvert}{\sqrt{p^{2}+q^{2}}}= \frac{1}{\sqrt{\left(\dfrac{1}{AB}\right)^{2}+\left(\dfrac{1}{AC}\right)^{2}}}.$$

Rearranging this relation gives

$$\sqrt{\left(\dfrac{1}{AB}\right)^{2}+\left(\dfrac{1}{AC}\right)^{2}}=\dfrac{1}{AD}.$$

Hence

$$\boxed{\lvert\mathbf{R}\rvert=\dfrac{1}{AD}}.$$

Therefore the correct option is
Option D which is: $$\dfrac{1}{AD}$$.