Let the values of p, for which the shortest distance between the lines $$\frac{x+1}{3} = \frac{y}{4} = \frac{z}{5}$$ and $$\vec{r} = (p\hat{i} + 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$$ is $$\frac{1}{\sqrt{6}}$$, be a, b (a < b). Then the length of the latus rectum of the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ is :

For the line $$\dfrac{x+1}{3}=\dfrac{y}{4}=\dfrac{z}{5}$$ put the common parameter $$t$$:
  $$x=-1+3t,\;y=0+4t,\;z=0+5t$$
Hence a point on the line is $$A(-1,\,0,\,0)$$ and its direction vector is $$\vec{d_1}= \langle 3,\,4,\,5\rangle$$.

The other line is given in vector form
  $$\vec{r}= (p\hat i+2\hat j+\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$$
so a point on it is $$B(p,\,2,\,1)$$ and its direction vector is $$\vec{d_2}= \langle 2,\,3,\,4\rangle$$.

Formula for shortest distance between two skew lines:
  $$D=\frac{\left|(\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2}))\right|}{\left|\vec{d_1}\times\vec{d_2}\right|}$$
where $$\overrightarrow{AB}=B-A=\langle p+1,\,2,\,1\rangle$$.

First evaluate the cross-product:
  $$\vec{d_1}\times\vec{d_2}= \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] 3 & 4 & 5\\ 2 & 3 & 4 \end{vmatrix} =\langle 1,\,-2,\,1\rangle$$
  $$|\vec{d_1}\times\vec{d_2}|=\sqrt{1^{2}+(-2)^{2}+1^{2}}=\sqrt6$$.

Triple scalar product:
  $$\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2}) =\langle p+1,\,2,\,1\rangle\cdot\langle 1,\,-2,\,1\rangle =(p+1)(1)+2(-2)+1(1)=p-2$$.

Given shortest distance $$D=\dfrac{1}{\sqrt6}$$, therefore
  $$\frac{|p-2|}{\sqrt6}=\frac{1}{\sqrt6}\;\;\Longrightarrow\;\;|p-2|=1$$
  $$\Rightarrow\;p=1\;\text{or}\;p=3$$.

Arranging $$a

The required ellipse is
  $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \;\Longrightarrow\; \frac{x^{2}}{1^{2}}+\frac{y^{2}}{3^{2}}=1$$
so the semi-major axis is $$b_{\max}=3$$ and the semi-minor axis is $$a_{\min}=1$$.

For an ellipse, length of the latus rectum $$L$$ is
  $$L=\frac{2\,(\,\text{(semi-minor)}\,)^{2}}{\text{(semi-major)}}$$
Hence
  $$L=\frac{2\,(1)^{2}}{3}=\frac{2}{3}$$.

Therefore the length of the latus rectum is $$\dfrac{2}{3}$$, which is Option C.