. Show that the lines with the equations

$\frac{x+1}{2}=\frac{y-3}{-4}=\frac{z-2}{5}\text{and}\frac{x-4}{3}=\frac{y+1}{2}=\frac{z}{3}$

are skew, find the equations of the parallel planes containing the lines, and find the distance between the lines.

The lines

$\frac{x+1}{2}=\frac{y-3}{-4}=\frac{z-2}{5}\text{and}\frac{x-4}{3}=\frac{y+1}{2}=\frac{z}{3}$

Consider the point $P(1,2,-3)$and the line with the equation $\mathbf{r}\left(t\right)=⟨2+5t,-3+t,-4t⟩$and the plane with the equation $x-3y+4z=5$

The direction vector of the line $\mathbf{r}\left(t\right)=⟨2+5t,-3+t,-4t⟩$is $\mathbf{d}=⟨5,1,-4⟩$$ and the normal vector of the plane $x-3y+4z=5$is $\mathbf{N}=⟨1,-3,4⟩$.

The dot product of direction vector and the normal vector is:

$\mathbf{d}·\mathbf{N}=⟨5,1,-4⟩·⟨1,-3,4⟩$

$=5-3-16=-14$

The dot product is non-zero. Therefore, line and the plane intersect.

The point of intersection is obtained for the value of parameter $t=t_0$such that the line $\mathbf{r}\left(t_0\right)$is on the plane.

The coordinates $\left(x\left(t_0\right),y\left(t_0\right),z\left(t_0\right)\right)$are given by:

$x\left(t_0\right)=2+5t_0$

$y\left(t_0\right)=-3+t_0$

$z\left(t_0\right)=-4t_0$

The points satisfy the plane $x-3y+4z=5.$

Therefore, on substitution the following equation is obtained.

$2+5t_0-3\left(-3+t_0\right)+4\left(-4t_0\right)=5\text{(Substitution)}$

$2+5t_0+9-3t_0-16t_0=5$

$-14t_0=-6$

$\Rightarrow t_0=\frac{6}{14}=\frac{3}{7}$

$\text{Therefore, point of intersection is:}$

$$\begin{gathered} x\left(t_0\right)=2+5\left(\frac{3}{7}\right) \\ =\frac{29}{7} \end{gathered}$$

$$\begin{gathered} y\left(t_0\right)=-3+\frac{3}{7} \\ =-\frac{18}{7} \end{gathered}$$

$$\begin{gathered} z\left(t_0\right)=-4\times \frac{3}{7} \\ =\frac{-12}{7} \end{gathered}$$

$\text{The point at which line with the equation}\mathbf{r}\left(t\right)=⟨2+5t,-3+t,-4t⟩\text{intersects the plane}$

$x-3y+4z=5\text{is}\left(\frac{29}{7},-\frac{18}{7},-\frac{12}{7}\right)$