There is no analytical solution, since that would include solving integral equations. There are however very fast numerical solutions. Those involve minizing distance function $d(t)=\left(\vec{r}(t)-\vec{p}\right)^2$ where $\vec{p}$ is the point you want to calculate distance to and $t$ parametrizes your clothoid which is given by $\vec{r}(t)$.
You have mentioned that your clothoids are "non-agressive" which I presume means that tangent from the start till the end of the clothoid does not rotate by a large angle. That will make $d(t)$ well behaved function of the parameter along the clothoid.
What you can do then is to subdivide your spline in sections where tangent at the beginning and the end make up angle not larger than 60 degrees. You are guaranteed that there will be no more than one extremum of $d(t)$ on that section. You can then check if $d'(t_0)d'(t_1) < 0$, where $t_0$ and $t_1$ are parameters of endpoints of your section. If this condition is satisfied you can apply for example Dekker's method to find a zero of $d'(t)=2\left(\vec{r}(t)-\vec{p}\right)\cdot\vec{r}'(t)$ (you can compute $\vec{r}'(t)$ analytically). It will converge in few iterations and you can then check whether you have minimum or maximum there by checking in final iteration whether $\left(d'(a_{k})-d'(b_{k})\right)/(a_k-b_k)>0$ (see Wikipedia article for the notation).
set the origin in A and then calculate the equation of the plane EBP finally calculate the distance of the plane from the origin A
if the plane equation is ax+by+cz=d the distance from the origin is given by
$$\frac {|d|}{\sqrt{a^2+b^2+c^2}}$$
Assuming the origin of the axis in A, AB=x-axis, AD=y-axis and AE= z-axis the equation of the plane EBP can be done at least in two ways:
1) by direct calculation imposing that E,B and P $\in%$ plane (you can assume wlog that d=1):
$$E=(0,0,3): a\cdot 0 + b \cdot 0 + c \cdot 3 = 1$$
$$B=(6,0,0): a\cdot 6 + b \cdot 0 + c \cdot 0 = 1$$
$$P=(3,4,0): a\cdot 3 + b \cdot 4 + c \cdot 0 = 1$$
you obtain the following linear system of three equation in three unknown:
$$3c= 1$$
$$6a= 1$$
$$3a+ 4b = 1$$
and then:$$c=1/3,a=1/6,b=1/8$$
finally the EBP plane equation:
$$\frac{1}{6}x+\frac{1}{8}y+\frac{1}{3}z=1$$
that is equivalent to:
$$4x+3y+8z=24$$
2) by cross product of two vectors BE and BP (but you can use another pair)
in this case you obtain: BE=(-6,0,3) and BP=(-3,4,0)
$$\begin{vmatrix}
i & j & k \\
-6 & 0 & 3 \\
-3 & 4 & 0
\end{vmatrix}$$
$$= -12 \cdot i- 9 \cdot j-24 \cdot k$$
that is a normal vector to the plane EBP which components coincide with the coefficients a,b,c of the plane EBP, thus the equation of the plane EPB is:
$$-12x-9y-24z=d$$
imposing that B $\in$ EPB:
$$-12 \cdot 6 = -72 = d$$
and finally:
$$-12x-9y-24z=-72$$
that is equivalent to (dividing both side by -3):
$$4x+3y+8z=24$$
once you have the plane equation the distance of the plane EPB from the origin A is given by:
$$\frac {|24|}{\sqrt{4^2+3^2+8^2}}=\frac {24}{\sqrt{89}}=\frac {24}{89}\sqrt{89}$$
Best Answer
As you say, we have 3 points, $A(10,10)$, $B(15,20)$, and $C(16,16)$.
The distance formula says that the distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is $$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$$
Here, the distance $AB$ is $5\sqrt5$
Similarly, $AC$ is $6\sqrt2$
And $BC$ is $\sqrt{17}$
Since the perpendicular is the smallest distance, let a point $D(x,y)$ on $AB$ such that $AB\underline{|} CD$, we can form 2 right triangles $\delta ADC$ and $\delta BDC$
We already know that $AB$ is $5\sqrt5$
Let $CD=x$
Applying the Pythagorean theorem:-
$AD=\sqrt{72-x^2}$
And
$BD=\sqrt{17-x^2}$
Since $AB=AD+BD$
Therefore
$$5\sqrt5=\sqrt{72-x^2}+\sqrt{17-x^2}$$
Solving, $x={6\over\sqrt5}$
Therefore, here the smallest distance is $6\over\sqrt5$