If $$\lambda \in R$$ is such that the sum of the cubes of the roots of the equation, $$x^2 + (2 - \lambda)x + (10 - \lambda) = 0$$ is minimum, then the magnitude of the difference of the roots of this equation is:

Let the two roots of the quadratic equation

$$x^{2} + (\,2-\lambda\,)x + (\,10-\lambda\,) = 0$$

be denoted by $$\alpha$$ and $$\beta$$. We first relate the elementary symmetric functions of the roots to the parameter $$\lambda$$.

From the standard relations for a monic quadratic $$x^{2}+Bx+C=0,$$ we have

$$\alpha+\beta = -B, \quad \alpha\beta = C.$$

Comparing, $$B = 2-\lambda$$ and $$C = 10-\lambda,$$ so

$$\alpha+\beta = - (\,2-\lambda\,) = \lambda-2, \qquad \alpha\beta = 10-\lambda.$$

Now we form the required expression, the sum of cubes of the roots. Using the identity

$$\alpha^{3} + \beta^{3} = (\alpha+\beta)^{3} \;-\; 3\alpha\beta(\alpha+\beta),$$

we substitute the expressions just found:

$$\alpha^{3} + \beta^{3} = (\lambda-2)^{3} \;-\; 3(10-\lambda)(\lambda-2).$$

To simplify manipulation, set

$$t = \lambda-2 \quad\Longrightarrow\quad \lambda = t+2,$$

which also gives

$$\alpha+\beta = t, \qquad \alpha\beta = 10-(t+2)=8-t.$$

Hence

$$\alpha^{3} + \beta^{3} \;=\; t^{3} \;-\; 3t(8-t).$$

Expanding the right-hand side term by term,

$$t^{3} \;-\; 3t(8-t) = t^{3} \;-\; 24t + 3t^{2},$$

so the function to be minimised is

$$S(t) = t^{3} + 3t^{2} - 24t.$$

We differentiate with respect to $$t$$:

$$\dfrac{dS}{dt} = 3t^{2} + 6t - 24.$$

Setting the derivative to zero for extrema,

$$3t^{2} + 6t - 24 = 0 \;\;\Longrightarrow\;\; t^{2} + 2t - 8 = 0.$$

Solving this quadratic,

$$t = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2} \implies t = 2 \quad\text{or}\quad t = -4.$$

To determine which value gives the minimum, compute the second derivative:

$$\dfrac{d^{2}S}{dt^{2}} = 6t + 6.$$

At $$t = 2$$ we get $$6(2)+6 = 18 > 0,$$ indicating a local minimum; at $$t = -4$$ we get $$6(-4)+6 = -18 < 0,$$ indicating a local maximum. Therefore the minimum value of $$\alpha^{3}+\beta^{3}$$ occurs at

$$t = 2 \quad\Longrightarrow\quad \lambda = t+2 = 4.$$

With this specific parameter value, the sum and product of the roots become

$$\alpha+\beta = \lambda-2 = 4-2 = 2, \qquad \alpha\beta = 10-\lambda = 10-4 = 6.$$

We now need the magnitude of the difference of the roots. Using the identity

$$(\alpha-\beta)^{2} = (\alpha+\beta)^{2} \;-\; 4\alpha\beta,$$

we substitute the just-obtained sum and product:

$$ (\alpha-\beta)^{2} = 2^{2} - 4\cdot 6 = 4 - 24 = -20.$$

Since this value is negative, the roots are complex conjugates; write them as

$$\alpha = u + iv,\;\; \beta = u - iv.$$

Then $$\alpha-\beta = 2iv,$$ and

$$|\,\alpha-\beta\,| = 2|v| = 2\sqrt{5},$$

because $$v^{2} = 5$$ (found from $$u^{2}+v^{2} = 6$$ and $$u = 1$$ which comes from $$\alpha+\beta = 2u = 2$$).

Thus the magnitude of the difference of the roots is

$$2\sqrt{5}.$$

Hence, the correct answer is Option B.