Rationalization. Geometrically speaking, two lines \(l_1\) and \(l_2\) are parallel if and only if the cross products of their directional vectors a and b is the zero vector \(<0,0,0>=\vec{0}\). That is:

\(a\times b=\vec{0}\)

Two lines are interesting if there exists a point (x,y,z) that is common in their domain. Lastly, two lines are skew if they are neither parallel or intersecting. We are given two sets of symmetric equations:

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

\(\frac{x-3}{1}=\frac{y+4}{3}=\frac{z-2}{7}\)

Two lines must be one of the three (parallel, intersecting, or skew).

Acquiring the directional vectors. The directional vectors of a parametric equation correspond to the denominators of each line. Thus, we get:

\(\frac{x-2}{1}=\frac{y-3}{-2}=\frac{z-1}{-3}\Rightarrow v_1=<1,-2,-3>\)

\(\frac{x-3}{1}=\frac{y+4}{3}=\frac{z-2}{-7}\Rightarrow v_2=<1,3,-7>\)

Checking if the vectors are parallel. We can check if the vectors are parallel through the cross product or by ratio and proportion. We do not need to do that here fortunately. Observe that the x-component of the vectors is the same but no the other components, which means that they are not parallel.

Checking if the vectors are intersecting: rationalization. A common way of doing this is by equating x and y and solving for parameters s and t. However, the given equations are still in symmetric form. Let's rewrite these symmetric equations into parametric form by equating to parameters s and t, respectively:

\(L_1:t=\frac{x-2}{1}=\frac{y-3}{-2}=\frac{z-1}{-3}\Rightarrow\begin{cases}x=t+2\\y=-2t+3\\z=-3t+1\end{cases}\)

\(L_2:s=\frac{x-3}{1}=\frac{y+4}{3}=\frac{z-2}{-7}\Rightarrow\begin{cases}x=s+3\\y=3s-4\\z=-7s+2\end{cases}\)

Substituting s and t to the parametric equations for z must yield an equality. We are checking the consistency of the linear system. Doing that, we get:

\(\begin{cases}x=s+3\\y=3s-4\\z=-7s+2\end{cases}=\begin{cases}x=t+2\\y=-2t+3\\z=-3t+1\end{cases}\)

\(t+2=s+3\)

\(=2t+3=3s-4\)

\(-3t+1=7s+2\)

Finding the intersection (x,y,z). To find the point of intersection (x,y,z), simply substitute the calculated values s=1 and t=2 to either equation

\(L_1:\begin{cases}x=2+2\\y=-2(2)+3\\z=-3(2)+1\end{cases}\Rightarrow(x,y,z)=(4,-1,-5)\)

Conclusion. There are values of s and t such that both equations are equal due to the consistency of the system. The system are intersecting at point (4,-1,-5)

\(a\times b=\vec{0}\)

Two lines are interesting if there exists a point (x,y,z) that is common in their domain. Lastly, two lines are skew if they are neither parallel or intersecting. We are given two sets of symmetric equations:

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

\(\frac{x-3}{1}=\frac{y+4}{3}=\frac{z-2}{7}\)

Two lines must be one of the three (parallel, intersecting, or skew).

Acquiring the directional vectors. The directional vectors of a parametric equation correspond to the denominators of each line. Thus, we get:

\(\frac{x-2}{1}=\frac{y-3}{-2}=\frac{z-1}{-3}\Rightarrow v_1=<1,-2,-3>\)

\(\frac{x-3}{1}=\frac{y+4}{3}=\frac{z-2}{-7}\Rightarrow v_2=<1,3,-7>\)

Checking if the vectors are parallel. We can check if the vectors are parallel through the cross product or by ratio and proportion. We do not need to do that here fortunately. Observe that the x-component of the vectors is the same but no the other components, which means that they are not parallel.

Checking if the vectors are intersecting: rationalization. A common way of doing this is by equating x and y and solving for parameters s and t. However, the given equations are still in symmetric form. Let's rewrite these symmetric equations into parametric form by equating to parameters s and t, respectively:

\(L_1:t=\frac{x-2}{1}=\frac{y-3}{-2}=\frac{z-1}{-3}\Rightarrow\begin{cases}x=t+2\\y=-2t+3\\z=-3t+1\end{cases}\)

\(L_2:s=\frac{x-3}{1}=\frac{y+4}{3}=\frac{z-2}{-7}\Rightarrow\begin{cases}x=s+3\\y=3s-4\\z=-7s+2\end{cases}\)

Substituting s and t to the parametric equations for z must yield an equality. We are checking the consistency of the linear system. Doing that, we get:

\(\begin{cases}x=s+3\\y=3s-4\\z=-7s+2\end{cases}=\begin{cases}x=t+2\\y=-2t+3\\z=-3t+1\end{cases}\)

\(t+2=s+3\)

\(=2t+3=3s-4\)

\(-3t+1=7s+2\)

Finding the intersection (x,y,z). To find the point of intersection (x,y,z), simply substitute the calculated values s=1 and t=2 to either equation

\(L_1:\begin{cases}x=2+2\\y=-2(2)+3\\z=-3(2)+1\end{cases}\Rightarrow(x,y,z)=(4,-1,-5)\)

Conclusion. There are values of s and t such that both equations are equal due to the consistency of the system. The system are intersecting at point (4,-1,-5)