If the tangent to the curve $$y = f(x) = x\log_e x$$, $$(x > 0)$$ at a point $$(c, f(c))$$ is parallel to the line-segment joining the points $$(1, 0)$$ and $$(e, e)$$, then $$c$$ is equal to:

We are given the curve $$y=f(x)=x\log_e x$$ with the condition $$x>0$$ because the natural logarithm is defined only for positive numbers. We look at the tangent to this curve at the general point $$(c,f(c))$$(that is, at $$x=c$$). The problem states that this tangent line is parallel to the straight line that joins the two fixed points $$(1,0)$$ and $$(e,e)$$. Our task is to find the value of $$c$$ for which this happens.

First we recall the basic fact from coordinate geometry: if two lines are parallel, then their slopes are equal. Therefore, we must equate

1. the slope of the tangent to the curve at $$x=c$$ and
2. the slope of the line segment connecting $$(1,0)$$ and $$(e,e)$$.

We begin by computing the slope of the tangent to the curve. For that we need the derivative of $$f(x)=x\log_e x$$. We use the standard differentiation rule:

Derivative rule (Product Rule): if $$u(x)$$ and $$v(x)$$ are differentiable, then $$\dfrac{d}{dx}\,[u(x)v(x)]=u'(x)v(x)+u(x)v'(x).$$

Here we identify $$u(x)=x$$ and $$v(x)=\log_e x$$, whose derivatives are $$u'(x)=1$$ and $$v'(x)=\dfrac{1}{x}$$ respectively. Applying the product rule gives

$$\frac{d}{dx}\,[x\log_e x]=1\cdot\log_e x + x\cdot\frac{1}{x}=$$\quad $$\log_e x + 1.$$

Thus the slope of the tangent to the curve $$y=f(x)$$ at an arbitrary point $$x=c$$ is

$$f'(c)=\log_e c + 1.$$

Now we find the slope of the line segment joining the two given points $$(1,0)$$ and $$(e,e)$$. The two-point slope formula from coordinate geometry states:

For points $$(x_1,y_1)$$ and $$(x_2,y_2)$$, the slope $$m$$ of the line passing through them is $$m=\dfrac{y_2-y_1}{\,x_2-x_1\,}.$$

Taking $$(x_1,y_1)=(1,0)$$ and $$(x_2,y_2)=(e,e)$$, we obtain

$$m=\frac{e-0}{\,e-1\,}=\frac{e}{\,e-1\,}.$$

Because the tangent line must be parallel to this segment, their slopes are equal. So we set

$$f'(c)=\frac{e}{\,e-1\,}.$$

Substituting the expression for $$f'(c)$$ we found earlier, we have

$$\log_e c + 1 = \frac{e}{\,e-1\,}.$$

We now solve this equation step by step for $$c$$. First isolate the logarithmic term on the left:

$$\log_e c = \frac{e}{\,e-1\,} - 1.$$

To combine the right-hand side into a single fraction, write $$1$$ with denominator $$(e-1)$$:

$$1 = \frac{e-1}{\,e-1\,}.$$

Hence

$$\log_e c = \frac{e}{\,e-1\,} - \frac{e-1}{\,e-1\,}.$$

Subtracting the numerators gives

$$\log_e c = \frac{\,e - (e-1)\,}{\,e-1\,} = \frac{e - e + 1}{\,e-1\,} = \frac{1}{\,e-1\,}.$$

We now remove the logarithm by using the definition of the natural logarithm: if $$\log_e c = k$$, then $$c = e^{\,k}$$. Applying this with $$k = \dfrac{1}{\,e-1\,}$$, we get

$$c = e^{\left(\frac{1}{e-1}\right)}.$$

This value exactly matches Option B in the given list.

Hence, the correct answer is Option B.