Answer
The lines are skew.
Work Step by Step
(a) The line in 2d space that passes through the point $P_{0}\left(x_{0}, y_{0}\right)$ and is parallel to the non-zero vector $\mathbf{v}=\langle a, b\rangle=a \mathbf{i}+b \mathbf{j}$ has parametric equations
\[
b t+y_{0}=y, \quad a t+x_{0}=x
\]
(b) The line in 3d space that passes through the point $P_{0}\left(x_{0}, y_{0}, z_{0}\right)$ and that is parallel to the non-zero vector $\mathbf{v}=\langle a, b, c\rangle=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}$ has the parametric equations
\[
c t+z_{0}=z, \quad b t+y_{0}=y, \quad at+x_{0}=x
\]
The vector equations of these lines can be written as:
\[
\begin{aligned}
\langle x, y, z\rangle &=\left\langle x_{0}+a t, y_{0}+b t, z_{0}+c t\right\rangle \\
\langle x, y\rangle &=\left\langle x_{0}+a t, y_{0}+b t\right\rangle
\end{aligned}
\]
or
\[
\begin{aligned}
\langle x, y, z\rangle &=\left\langle x_{0}, y_{0}, z_{0}\right\rangle+t\langle a, b, c\rangle \\
\langle x, y\rangle &=\left\langle x_{0}, y_{0}\right\rangle+t\langle a, b\rangle
\end{aligned}
\]
Since the line:
\[
-3 t+5=z, \quad t+3=y, \quad 7 t+1=x
\]
is parallel to $\vec{v}_{1}=\langle 7,1,-3\rangle$ and the line:
\[
2 t+7=z, -3 t+5=y, \quad -t+4=x
\]
is parallel to $\vec{v}_{2}=\langle-1,-3,2\rangle,$ and $\vec{v}_{1} \times \vec{v}_{2} \neq 0,$ these lines are skew. We notice that if $\vec{v}_{1} / / \vec{v}_{2},$ then $L_{1} / / L_{2}$ and $\vec{v}_{1} \times \vec{v}_{2}=0 .$ In this case
\[
\vec{v}_{1} \times \vec{v}_{2}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
7 & 1 & -3 \\
-1 & -3 & 2
\end{array}\right|=(2-9) \hat{i}+(14-3) \hat{j}+(-21+1) \hat{k} \neq 0
\]
So, the given lines are skew.
