Step 1: Setting up parametric coordinates:
Line 1: \((1+2\lambda,\ 1-\lambda,\ \lambda)\). Line 2: \((2+3\mu,\ 1-5\mu,\ -1+2\mu)\).
Step 2: Forming the connecting vector and perpendicularity conditions:
The vector between a general point on each line is \(\vec V=(1+3\mu-2\lambda,\ -5\mu+\lambda,\ -1+2\mu-\lambda)\); for the shortest connector, \(\vec V\) must be perpendicular to both direction vectors — this reduces to solving the same determinant/cross-product system as the standard formula.
Step 3: Applying the compact vector formula:
Rather than solving the two perpendicularity equations directly, use \(d=\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}\), which is derived from exactly that perpendicularity condition and gives \(d=10/\sqrt{59}\) directly.
Final Answer:
\[ \boxed{\dfrac{10}{\sqrt{59}}\text{ units}} \]