If the shortest distance between the straight lines $$3(x-1) = 6(y-2) = 2(z-1)$$ and $$4(x-2) = 2(y-\lambda) = (z-3)$$, $$\lambda \in R$$ is $$\frac{1}{\sqrt{38}}$$, then the integral value of $$\lambda$$ is equal to:

Rewrite the two lines in symmetric form. For Line 1, divide $$3(x-1) = 6(y-2) = 2(z-1)$$ by 6 to get $$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$$, so the line passes through $$A = (1, 2, 1)$$ with direction vector $$\vec{b_1} = (2, 1, 3)$$.

For Line 2, set $$4(x-2) = 2(y-\lambda) = (z-3) = t$$. Then $$x = 2 + \frac{t}{4}$$, $$y = \lambda + \frac{t}{2}$$, $$z = 3 + t$$. Reading off the coefficients of $$t$$, the direction vector is proportional to $$\left(\frac{1}{4}, \frac{1}{2}, 1\right)$$, or equivalently (multiplying by 4) $$\vec{b_2} = (1, 2, 4)$$. The line passes through $$B = (2, \lambda, 3)$$.

Compute the cross product: $$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix} = (1 \cdot 4 - 3 \cdot 2)\hat{i} - (2 \cdot 4 - 3 \cdot 1)\hat{j} + (2 \cdot 2 - 1 \cdot 1)\hat{k} = -2\hat{i} - 5\hat{j} + 3\hat{k}$$. Its magnitude is $$|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 25 + 9} = \sqrt{38}$$.

The vector joining the two points is $$\overrightarrow{AB} = B - A = (2 - 1,\; \lambda - 2,\; 3 - 1) = (1,\; \lambda - 2,\; 2)$$.

The shortest distance formula for skew lines gives $$d = \frac{|\overrightarrow{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$. The dot product is $$(1)(-2) + (\lambda - 2)(-5) + (2)(3) = -2 - 5\lambda + 10 + 6 = 14 - 5\lambda$$.

Setting $$d = \frac{1}{\sqrt{38}}$$: $$\frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}}$$, giving $$|14 - 5\lambda| = 1$$. This yields $$14 - 5\lambda = 1$$ (so $$\lambda = \frac{13}{5}$$) or $$14 - 5\lambda = -1$$ (so $$\lambda = 3$$).

Since the problem asks for the integral value of $$\lambda$$, the answer is $$\lambda = 3$$.