The direction vectors are the numbers in the denominators of the symmetric equations.

\(\displaystyle{L}_{{1}}:{<}{1},-{1},{3}{>}\)

\(\displaystyle{L}_{{2}}:{<}{2},-{2},{7}{>}\)

One vector is not a multiple of the other. 2 times 1 is 2, but 2 times 3 isn't 7. The lines are not parallel.

Parametric form of \(\displaystyle{L}_{{1}}\)

\(\displaystyle{\frac{{{x}}}{{{1}}}}={t}\Rightarrow{x}={t}\)

\(\displaystyle{\frac{{{y}-{1}}}{{-{1}}}}={t}\Rightarrow{y}={1}-{t}\)

\(\displaystyle{\frac{{{z}-{2}}}{{{3}}}}={t}\Rightarrow{z}={2}+{3}{t}\)

Parametric form of \(\displaystyle{L}_{{2}}\)

\(\displaystyle{\frac{{{x}-{2}}}{{{2}}}}={s}\Rightarrow{x}={2}+{2}{s}\)

\(\displaystyle{\frac{{{y}-{3}}}{{-{2}}}}={s}\Rightarrow{y}={3}-{2}{s}\)

\(\displaystyle{\frac{{{z}}}{{{7}}}}={s}\Rightarrow{z}={7}{s}\)

See if the lines intersect by setting the component parts equal to each other and seeing if there is a solution.

\(\displaystyle{t}={2}+{2}{s}\)

\(\displaystyle{1}-{t}={3}-{2}{s}\)

\(\displaystyle{2}+{3}{t}={7}{s}\)

Substitute for t

\(\displaystyle{1}-{\left({2}+{2}{s}\right)}={3}-{2}{s}\)

\(\displaystyle{1}-{2}-{2}{s}={3}-{2}{s}\)

\(\displaystyle-{1}={3}\)

This is a contradiction, so there is not solution. The lines do not intersect so they are skew.

\(\displaystyle{L}_{{1}}:{<}{1},-{1},{3}{>}\)

\(\displaystyle{L}_{{2}}:{<}{2},-{2},{7}{>}\)

One vector is not a multiple of the other. 2 times 1 is 2, but 2 times 3 isn't 7. The lines are not parallel.

Parametric form of \(\displaystyle{L}_{{1}}\)

\(\displaystyle{\frac{{{x}}}{{{1}}}}={t}\Rightarrow{x}={t}\)

\(\displaystyle{\frac{{{y}-{1}}}{{-{1}}}}={t}\Rightarrow{y}={1}-{t}\)

\(\displaystyle{\frac{{{z}-{2}}}{{{3}}}}={t}\Rightarrow{z}={2}+{3}{t}\)

Parametric form of \(\displaystyle{L}_{{2}}\)

\(\displaystyle{\frac{{{x}-{2}}}{{{2}}}}={s}\Rightarrow{x}={2}+{2}{s}\)

\(\displaystyle{\frac{{{y}-{3}}}{{-{2}}}}={s}\Rightarrow{y}={3}-{2}{s}\)

\(\displaystyle{\frac{{{z}}}{{{7}}}}={s}\Rightarrow{z}={7}{s}\)

See if the lines intersect by setting the component parts equal to each other and seeing if there is a solution.

\(\displaystyle{t}={2}+{2}{s}\)

\(\displaystyle{1}-{t}={3}-{2}{s}\)

\(\displaystyle{2}+{3}{t}={7}{s}\)

Substitute for t

\(\displaystyle{1}-{\left({2}+{2}{s}\right)}={3}-{2}{s}\)

\(\displaystyle{1}-{2}-{2}{s}={3}-{2}{s}\)

\(\displaystyle-{1}={3}\)

This is a contradiction, so there is not solution. The lines do not intersect so they are skew.