Example 1.28. Drop a perpendicular from the origin to the line

$$
\frac{x-2}{2}=\frac{y-1}{3}=\frac{z-3}{1} \text {. }
$$
Solution. Let's form the equation of the plane passing through the origin and perpendicular to the given line: $2 x+3 y+z=0$. (For this plane, we can take $A=l ; B=m$; $C=n ; D=0 ;$ the condition of perpendicularity of the line and the plane is used, see point 8 of the introduction to this section).

Let's find the point of intersection of this plane and the given line. The parametric equations of the line are:

$$
\begin{gathered}
x=2 t+2 \\
y=3 t+1 \\
z=t+3
\end{gathered}
$$

To determine $t$, we have the equation:

$$
2(2 t+2)+3(3 t+1)+t+3=0 .
$$

Thus, $t=-\frac{5}{7}$. The coordinates of the intersection point will be: $x=\frac{4}{7}, y=-\frac{8}{7}, z=\frac{16}{7}$, i.e., $M \frac{4}{7} ;-\frac{8}{7} ; \frac{16}{7}$.

It remains to form the equations of the line passing through the origin and through point $M$ (see point 3 of the introduction to this section):

$$
\frac{x}{\frac{4}{7}}=\frac{y}{-\frac{8}{7}}=\frac{z}{\frac{16}{7}} \quad \text { or } \quad \frac{x}{1}=\frac{y}{-2}=\frac{z}{4}
$$