4.6. A line is given by its general equations:

$$
\left\{\begin{array}{l}
3 x-4 y-z-1=0 \\
x+2 y+4 z-5=0
\end{array}\right.
$$

From these equations, obtain: 1) the canonical equations; 2) the parametric equations; 3) find the distance from the point $A(3,2 ; 4,4 ;-1)$ to the given line.
Solution.

1) To find the coordinates of two points belonging to the line, we first set $z_{0}=0$ and substitute it into the equation of the line:

$$
\left\{\begin{array} { l } 
{ 3 x _ { 0 } - 4 y _ { 0 } - 0 - 1 = 0 , } \\
{ x _ { 0 } + 2 y _ { 0 } + 4 \cdot 0 - 5 = 0 , }
\end{array} \Rightarrow \left\{\begin{array}{l}
3 x_{0}-4 y_{0}=1, \\
x_{0}+2 y_{0}=5,
\end{array} \Rightarrow x_{0}=2.2 ; \quad y_{0}=1.4\right.\right.
$$

The coordinates of the first point on the line are $P_{0}(2.2 ; 1.4 ; 0)$.

Now let's set, for example, $z_{1}=1$, and similarly find the coordinates of the second point on the line $P_{1}\left(x_{1}, y_{1}, z_{1}\right)$:

$$
\left\{\begin{array} { l } 
{ 3 x _ { 1 } - 4 y _ { 1 } - 1 - 1 = 0 , } \\
{ x _ { 1 } + 2 y _ { 1 } + 4 \cdot 1 - 5 = 0 , }
\end{array} \Rightarrow \left\{\begin{array}{l}
3 x_{1}-4 y_{1}=2, \\
x_{1}+2 y_{1}=1,
\end{array} \Rightarrow x_{1}=0.8 ; \quad y_{1}=0.1\right.\right.
$$

then the coordinates of the point $P_{1}(0.8 ; 0.1 ; 1)$.

We will form the coordinates of the current vector of the line

$$
\overrightarrow{P_{0} M}=(x-2.2 ; y-1.4 ; z-0)
$$

and the vector $\overrightarrow{P_{0} P_{1}}=(-1.4 ;-1.3 ; 1)$, belonging to the line. We write the condition of collinearity of vectors $\overline{P_{0} M}$ and $\overline{P_{0} P_{1}}$:

$$
(x-2.2) /(-1.4)=(y-1.4) /(-1.3)=z / 1
$$

Multiplying the last equations by ( $-0.1$ ), we get the canonical equations of the line in the following form:

$$
(x-2.2) / 14=(y-1.4) / 13=z /(-10)
$$

2) To find the parametric equations of the line, we set

$$
\begin{gathered}
(x-2.2) / 14=(y-1.4) / 13=z /(-10)=t \\
x=2.2+14 t, y=1.4+13 t, z=-10 t
\end{gathered}
$$

3) The distance from point $A$ to the line will be found using formula (4.16). In this case,

$$
\begin{gathered}
x_{1}=3.2, \quad y_{1}=4.4, z_{1}=-1, \\
x_{0}=2.2, y_{0}=1.4, z_{0}=0, \\
l=14, m=13, n=-10, \\
d=\frac{\sqrt{\left|\begin{array}{cc}
4.4-1.4 & -1-0 \\
13 & -10
\end{array}\right|^{2}+\left|\begin{array}{cc}
-1-0 & 3.2-2.2 \\
-10 & 14
\end{array}\right|^{2}+\left|\begin{array}{cc}
3.2-2.2 & 4.4-1.4 \\
14 & 13
\end{array}\right|^{2}}}{\sqrt{14^{2}+13^{2}+(-10)^{2}}} \\
=\frac{\sqrt{1146}}{\sqrt{465}} \cong 1.57 .
\end{gathered}
$$