Example 7. Find the equation of the plane containing the point $M_{1}(3,1,0)$ and the line $\frac{x-4}{1}=\frac{y}{2}=\frac{z-1}{3}$.
Solution. From the equation of the line, the coordinates of the point $M_{0}(4,0,1)$ on it and the directing vector $\vec{l}=\{1,2,3\}$ are known.

Let $M(x, y, z) \rightarrow$ the current point of the plane (Fig. 4.15). Then the vectors $\overrightarrow{M_{0} M}, \overrightarrow{M_{0} M_{1}}, \vec{l}$ lie in the same plane, i.e., they are coplanar. The condition of coplanarity of vectors (Chapter III, § 4) will be the desired equation: $\overrightarrow{M_{0} M} \cdot \overrightarrow{M_{0} M_{\mathrm{I}}} \cdot \vec{l}=0$. We have: $\overrightarrow{M_{0} M}=\{x-4, y, z-1\}$, $\overrightarrow{M_{0} M_{1}}=\{-1,1,-1\}$. The equation of the desired plane is

$$
\left|\begin{array}{ccc}
x-4 & y & z-1 \\
-1 & 1 & -1 \\
1 & 2 & 3
\end{array}\right|=0
$$

Expanding the determinant by the elements of the first row, we simplify: $5 x+2 y-3 z-17=0$

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-048.jpg?height=389&width=395&top_left_y=994&top_left_x=155)

Fig. 4.15

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-048.jpg?height=346&width=457&top_left_y=1042&top_left_x=699)

Fig. 4.16