Step 1: Set up the same formula but compute the cross product using cofactor expansion along a different row, as a cross-check:
Expand $\vec d_1\times\vec d_2$ along the $\hat k$ row logic directly: $\hat k$-component $=2(-5)-1(3)=-13$; $\hat j$-component (remember the sign flip) $=-(2\cdot2-1\cdot3)=-1$; $\hat i$-component $=1\cdot2-1\cdot(-5)=7$. This gives the same $7\hat i-\hat j-13\hat k$.
Step 2: Compute the connecting vector between a point on each line:
Point on line 1: $(1,1,0)$. Point on line 2: $(2,1,1)$. Connecting vector $=(2-1,1-1,1-0)=(1,0,1)$.
Step 3: Take the scalar triple product and divide by the magnitude of the cross product:
$(1,0,1)\cdot(7,-1,-13)=7+0-13=-6$; magnitude of cross product $=\sqrt{7^2+1^2+13^2}=\sqrt{219}$.
Step 4: Take the absolute value for the distance (distance is never negative):
Distance $=\dfrac{|-6|}{\sqrt{219}}=\dfrac{6}{\sqrt{219}}$.
Final Answer:
Shortest distance $=\dfrac{6}{\sqrt{219}}$ units.
\[ \boxed{\dfrac{6}{\sqrt{219}}} \]