Example 1.26. Reduce to canonical form the equations of the line $2 x-y+3 z-1=0$ and $5 x+4 y-z-7=0$.
Solution. Excluding $y$ first, and then $z$, we get:

$$
13 x+11 z-11=0 \quad \text { and } \quad 17 x+11 y-22=0 .
$$

If we solve each of the equations for $x$, we will have:

$$
x=\frac{11(y-2)}{-17}=\frac{11(z-1)}{-13}
$$

from which

$$
\frac{x}{-11}=\frac{y-2}{17}=\frac{z-1}{13} \text {. }
$$

Second method: find the vector $\bar{S}=l \bar{i}+m \bar{j}+n \bar{k}$, parallel to the desired line. Since it must be perpendicular to the normal vectors of the given planes $\bar{N}_{1}=2 \bar{i}-\bar{j}+3 \bar{k}$ and $\bar{N}_{2}=5 \bar{i}+4 \bar{j}-\bar{k}$, we can take the vector product of vectors $\bar{N}_{1}$ and $\bar{N}_{2}$ as it.

$$
\bar{S}=\bar{N}_{1} \times \bar{N}_{2}=\left|\begin{array}{rrr}
\bar{i} & \bar{j} & \bar{k} \\
2 & -1 & 3 \\
5 & 4 & -1
\end{array}\right|=-11 \bar{i}+17 \bar{j}+13 \bar{k}
$$

Thus, $l=-11 ; m=17 ; n=13$.

We can take the point $M_{1}\left(x_{1}, y_{1}, z_{1}\right)$, through which the desired line passes, as the point of intersection of the line with any of the coordinate planes, for example, with the plane $y o z$. Since in this case $x_{1}=0$, the coordinates $y_{1}$ and $z_{1}$ of this point will be determined from the system of equations of the given planes, if we set $x=0$ in them:

$$
\begin{gathered}
-y+3 z-1=0 \\
4 y-z-7=0
\end{gathered}
$$

Solving this system, we find $y_{1}=2 ; z_{1}=1$.

Thus, the desired line is defined by the equations:

$$
\frac{x}{-11}=\frac{y-2}{17}=\frac{z-1}{13} .
$$

We have obtained the same answer.