42. Point $M$ in space is defined by the radius vector $\vec{a}=\overrightarrow{O M}$ with coordinates ( $x_{1}, y_{1}, z_{1}$ ). A line passing through point $M$ is defined by the unit vector $\vec{e}$, which forms angles $\alpha, \beta, \gamma$ with the axes $O x, O y$, $O z$. Let $\vec{r}$ be the variable radius vector corresponding to the points on the considered line. Given that the vector $\vec{r}-\vec{a}$ is collinear with the vector $\vec{e}$, write the equation of the line passing through the given point in vector form, and then - the system of equations in coordinates.
42. The vector $\vec{r}-\vec{a}$ is collinear with the vector $\vec{e}$, hence: $(\vec{r}-\vec{a})=t \vec{e}$, where $t \in \boldsymbol{R}$. Transitioning to coordinates, we find:

$$
\left(x-x_{1}\right) \vec{i}+\left(y-y_{1}\right) \vec{j}+\left(z-z_{1}\right) \vec{k}=t \vec{i} \cos \alpha+t \vec{j} \cos \beta+t \vec{k} \cos \gamma
$$

or

$$
\begin{gathered}
x-x_{1}=t \cos \alpha, y-y_{1}=t \cos \beta, z-z_{1}=t \cos \gamma \\
\frac{x-x_{1}}{\cos \alpha}=\frac{y-y_{1}}{\cos \beta}=\frac{z-z_{1}}{\cos \gamma}
\end{gathered}
$$

These are the canonical equations of a line passing through a given point in a given direction.