I have the straight lines:
$$d_1: \frac{x-1}2=\frac{y-3}1=\frac{z+2}1\\[4ex] d_2: \dfrac{x-1}1=\frac{y+2}{-4}=\frac{z-9}2$$
And I have to find the common perpendicular of these lines.
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$\begingroup$In general, let $\mathcal{H}$ be a Hilbert space over $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$, in which $\langle\_,\_\rangle$ is the inner product where the first entry is linear over $\mathbb{K}$ and the second entry is antilinear over $\mathbb{K}$ (and $\|\_\|$ is the norm associated to this inner product). A line in $\mathcal{H}$ is a set of the form $\ell(\textbf{x},\textbf{a}):=\{\mathbf{x}t+\mathbf{a}\,|\,t\in\mathbb{K}\}$ for fixed $\mathbf{x},\mathbf{a}\in\mathcal{H}$ such that $\mathbf{x}$ is nonzero.
Suppose we have two lines $l_1:=\ell\left(\textbf{x}_1,\textbf{a}_1\right)$ and $l_2:=\ell\left(\textbf{x}_2,\textbf{a}_2\right)$ in general position (that is, $\textbf{x}_1$ and $\textbf{x}_2$ are not proportional). Without loss of generality, assume that $\left\|\textbf{x}_i\right\|=1$ for both $i=1$ and $i=2$. Hence, $\epsilon:=\left\langle \textbf{x}_1,\textbf{x}_2\right\rangle$ satisfies $|\epsilon|<1$ due to the Cauchy-Schwarz Inequality and the assumption that the lines are in general position.
We shall prove that there exist unique points $P_i\in l_i$ for $i\in\{1,2\}$ such that $\underline{P_1P_2}:=P_2-P_1$ is perpendicular to both $l_1$ and $l_2$. That is, there is a unique common perpendicular line for $l_1$ and $l_2$, provided that $l_1$ does not intersect $l_2$. (Note that this statement includes the degenerate case where $P_1=P_2$, i.e., when $l_1$ intersects $l_2$. The zero element is orthogonal to any element of $\mathcal{H}$ anyhow.)
If $P_i=\textbf{x}_it_i+\textbf{a}_i$ for some $t_i\in\mathbb{K}$, where $i\in\{1,2\}$, then the conditions $\underline{P_1P_2}\perp l_i$ for $i=1$ and for $i=2$ is equivalent to demanding that $$\left\langle \textbf{x}_1t_1+\textbf{a}_1-\textbf{x}_2t_2-\textbf{a}_2,\textbf{x}_1\right\rangle=0\text{ and }\left\langle \textbf{x}_1t_1+\textbf{a}_1-\textbf{x}_2t_2-\textbf{a}_2,\textbf{x}_2\right\rangle=0\,.$$ Therefore, $$t_1-\bar\epsilon t_2=-\left\langle \textbf{a}_1-\textbf{a}_2,\textbf{x}_1\right\rangle\text{ and }\epsilon t_1-t_2=-\left\langle\textbf{a}_1-\textbf{a}_2,\textbf{x}_2\right\rangle\,,$$ whence $$t_1=-\frac{\left\langle\textbf{a}_1-\textbf{a}_2,\textbf{x}_1-\epsilon\textbf{x}_2\right\rangle}{1-|\epsilon|^2}\text{ and }t_2=-\frac{\left\langle\textbf{a}_2-\textbf{a}_1,\textbf{x}_2-\bar{\epsilon}\textbf{x}_1\right\rangle}{1-|\epsilon|^2}\,.$$ Without the unit norm assumption, $$t_1=-\frac{ \Big\langle\textbf{a}_1-\textbf{a}_2,\left\|\textbf{x}_2\right\|^2\textbf{x}_1-\left\langle\textbf{x}_1,\textbf{x}_2\right\rangle\textbf{x}_2\Big\rangle}{\left\|\textbf{x}_1\right\|^2\left\|\textbf{x}_2\right\|^2-\big|\left\langle \textbf{x}_1,\textbf{x}_2\right\rangle\big|^2}$$ and $$t_2=-\frac{\Big\langle\textbf{a}_2-\textbf{a}_1,\left\|\textbf{x}_1\right\|^2\textbf{x}_2-\left\langle\textbf{x}_2,\textbf{x}_1\right\rangle\textbf{x}_1\Big\rangle}{\left\|\textbf{x}_2\right\|^2\left\|\textbf{x}_1\right\|^2-\big|\left\langle \textbf{x}_2,\textbf{x}_1\right\rangle\big|^2}\,.$$ Hence, $P_1$ and $P_2$ exist and are unique, as they correspond to $t_1$ and $t_2$, which have been proven to uniquely exist.
Now, for the OP's particular problem, $\mathbb{K}=\mathbb{R}$, $\mathcal{H}=\mathbb{R}^3$ with the standard inner product, $\textbf{x}_1=\frac{(2,1,1)}{\sqrt{6}}$, $\textbf{x}_2=\frac{(1,-4,2)}{\sqrt{21}}$, $\textbf{a}_1=(1,3,-2)$, and $\textbf{a}_2=(1,-2,9)$. Ergo, $\epsilon=\bar{\epsilon}=0$, which leads to $t_1=\sqrt{6}$ and $t_2=-2\sqrt{21}$. Thus, $$P_1=(2,1,1)+(1,3,-2)=(3,4,-1)$$ and $$P_2=(-2,8,-4)+(1,-2,9)=(-1,6,5)\,.$$
$\endgroup$ $\begingroup$The classical method is as follows
First convert line equations into parametrical form
$$M_s(x=x_1+u_1s, \ \ y=y_1+v_1s, \ \ z=z_1+w_1s) \ \ \ (1)$$
and
$$N_t(x=x_2+u_2t, \ \ y=y_2+v_2t, \ \ z=z_1+w_2t) \ \ (2)$$
(you have already directing vectors $(u_1,v_1,w_1)$ and $(u_2,v_2,w_2)$.
Take arbitrary points on your lines for $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$).
Then express that the distance, or more exactly the squared distance
$$(M_sN_t)^2=(x_2+u_2t-x_1-u_1s)^2+(y_2+v_2t-y_1-v_1s)^2+(z_1+w_2t-z_1-w_1s)^2$$
has a minimal value.
As this function of $s$ and $t$ has to be minimized, the necessary (in fact sufficient) condition is that the partial derivatives with respect to $s$ and with respect to $t$ of this expression are zero. You will obtain a linear system. Solve it, finding a certain $s_0$ and a certain $t_0$.
Now plug in (1) the value of $s=s_0$; you obtain a point $M_{s_0}$ common to the first line and thecommon perpendicular (P). The same for the second line giving a point $N_{t_0}$ belonging this time to the second line and (P). You know completely line (P) because you know two of its points. From there it is easy to find its equation.
$\endgroup$ $\begingroup$Convert the line equations to ax + by + c format. And then take cross product of the vectors represented by (a, b, c)
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