Let us name (0,1,2) as the point A

Point B is on the given line such that AB is perpendicular to the given line

Vector along the given line \(\displaystyle{r}_{{1}}={<}{1},-{1},{2}{>}\)

Vector along the given line A and B is

\(\displaystyle{r}_{{2}}={<}{1}+{t},{1}-{t},{2}{t}\succ{<}{0},{1},{2}\ge{<}{1}+{t},-{t},{2}{t}-{2}{>}\)

\(\displaystyle{r}_{{1}}\) and \(\displaystyle{r}_{{2}}\) must be perpendicular, hence their dot product must be 0

\(\displaystyle{<}{1},-{1},{2}{>}\cdot{<}{1}+{t},-{t},{2}{t}-{2}\ge{0}\)

\(\displaystyle{\left({1}+{t}\right)}+{t}+{\left({4}{t}-{4}\right)}={0}\)

\(\displaystyle{6}{t}-{3}={0}\)

\(\displaystyle{6}{t}={3}\)

\(\displaystyle{t}={0.5}\)

The required line passes through (0,1,2)

Required line is parallel to \(\displaystyle{r}_{{2}}={<}{1}+{0.5},-{0.5},{2}\cdot{0.5}-{2}\ge{<}{1.5},-{0.5},-{1}{>}\)

Recall that: Vector equation of a line passing a point with position vector a and parallel to the vector b is \(\displaystyle{r}{\left({t}\right)}={a}+{t}{b}\)

Therefore equation of the required line is

\(\displaystyle{r}{\left({t}\right)}={<}{0},{1},{2}{>}+{t}{<}{1.5},-{0.5},-{1}{>}\)

\(\displaystyle{r}{\left({t}\right)}={<}{1.5}{t},{1}-{0.5}{t},{2}-{t}{>}\)

Parametric equations are

\(\displaystyle{x}={1.5}{t},\ {y}={1}-{0.5}{t},\ {z}={2}-{t}\)

Result: \(\displaystyle{x}={1.5}{t},\ {y}={1}-{0.5}{t},\ {z}={2}-{t}\)