Example 1.29. In the equations of the line $\frac{x}{2}=\frac{y}{-3}=\frac{z}{n}$, determine the parameter $n$ so that this line intersects with the line $\frac{x+1}{3}=\frac{y+5}{2}=\frac{z}{1}$, and find the point of their intersection.
Solution. To find the parameter $n$, we use the condition of the intersection of 2 straight lines:

$$
\left|\begin{array}{ccc}
x_{1}-x_{0} & y_{1}-y_{0} & z_{1}-z_{0} \\
l_{1} & m_{1} & n_{1} \\
l_{2} & m_{2} & n_{2}
\end{array}\right|=0
$$

From this, it follows:

$$
\left|\begin{array}{rrr}
1 & 5 & 0 \\
3 & 2 & 1 \\
2 & -3 & n
\end{array}\right|=0
$$

or

$$
2 n+10+3-15 n=0, \quad 13 n=13, \quad n=1 .
$$

Therefore, the equations of the intersecting lines are: the required:

$$
\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}
$$

and the given:

$$
\frac{x+1}{3}=\frac{y+5}{2}=\frac{z}{1}
$$

To find the coordinates of the intersection point of these lines, express $x$ and $y$ in terms of $z$ from the first equation: $x=2 z, y=-3 z$. Substituting these values into the equality $\frac{x+1}{3}=\frac{y+5}{2}$, we have $\frac{2 z+1}{3}=\frac{-3 z+5}{2}$, from which $z=1$. Knowing $z$, we find $x$ and $y: x=2 z=2, y=-3 z=-3$.

Therefore, $M(2 ;-3 ; 1)$.