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For nonparallel lines, $\mathbf{n}={\mathbf{m}}_1\times {\mathbf{m}}_2$ is perpendicular to both direction vectors. Every displacement between the lines differs from ${\mathbf{a}}_2-{\mathbf{a}}_1$ by a vector in their span, so its component along $\mathbf{n}$ is fixed. Varying the two parameters cancels the component in that span. The shortest distance between skew lines is thereforeThis also gives zero for intersecting nonparallel lines. For parallel lines the denominator vanishes and the correct separate formula is $|({\mathbf{a}}_2-{\mathbf{a}}_1)\times {\mathbf{m}}_1|/|{\mathbf{m}}_1|$.

The given lines have vector formsHere ${\mathbf{m}}_1\times {\mathbf{m}}_2=(-3,0,3)$ and ${\mathbf{a}}_2-{\mathbf{a}}_1=(-4,-4,2)$. Their scalar triple product is $18$, givingAs a constructive check, the closest parameters are $t=2/3$ and $u=7/3$, where the connecting vector is $(3,0,-3)$ and is perpendicular to both lines.

The real logarithm requires $x>0$. The identically zero function is a solution and must be retained before dividing by $y$. For a nonzero solution, divide by $x$ and make the Bernoulli differential equation substitution $u=1/y$. This givesIntegration yields $u/x=-x\ln x+x+C$. Consequentlywith each nonzero branch restricted to an interval where its denominator is nonzero. The right-hand side of the first-order equation is locally Lipschitz in $y$ for $x>0$, so a nonzero solution cannot cross the zero solution at a finite positive $x$.

For $x\ne 0$, divide by $x^2$ to obtain the linear ordinary differential equation $y^{\prime }+3y/x=x^{-2}$. An integrating factor is $x^3$ on either interval not containing zero. Thus $(x^{3}y{)}^{\prime }=x$ andSubstituting gives $x^2{y}^{\prime }+3xy=1$ directly. No classical solution can include $x=0$, since the original equation there would assert $0=1$; constants on the two separate intervals need not agree.