If the shortest distance between the straight lines \( 3(x-1)=6(y-2)=2(z-1) \) and \( 4(x-2)=2(y-\lambda)=(z-3), \lambda \in \mathbb{R} \) is \( \dfrac{1}{\sqrt{38}} \), then the integral value of \( \lambda \) is equal to:
\( -1 \)
\( 5 \)
\( 3 \)
\( 2 \)