Given:

\(\displaystyle{v}_{{1}}={\left({0},{1},−{2}\right)}+{t}{\left({2},{3},−{1}\right)}\)

\(\displaystyle{v}_{{2}}={\left({2},−{1},{0}\right)}+{t}{\left({2},{3},−{1}\right)}\)

Let \(\displaystyle{A}={\left({0},{1},-{2}\right)}\), \(\displaystyle{B}={\left({2},-{1},{0}\right)}\)

Thus, the plane containing these two lines will contain A and B. Hence, the vector \(\displaystyle{A}{B}={\left({2},-{2},{2}\right)}\)

Normal to the plave shall be the cross product of AB and the direction ratio of the lines.

Thus, \(\displaystyle\vec{{{n}}}=\vec{{{A}{B}}}\times{d}{i}{r}{e}{c}{t}{i}{o}{n}\ {r}{a}{t}{i}{o}\ {o}{f}\ {t}{h}{e}\ {l}in{e}\)

\(\displaystyle\vec{{{n}}}={\left({2},-{2},{2}\right)}\times{\left({2},{3},-{1}\right)}\)

\[\vec{n}=\begin{bmatrix}i & j & k \\2 & -2 & 2 \\2 & 3 & -1 \end{bmatrix}=-4i+6j+10k\]

Thus, the equation of the plane can be written as

\(\displaystyle{\left({x}-{o}\right)}{i}+{\left({y}-{1}\right)}{j}+{\left({z}+{2}\right)}\times\vec{{{n}}}={0}\)

\(\displaystyle-{4}{\left({x}-{0}\right)}+{6}{\left({y}-{1}\right)}+{10}{\left({z}+{2}\right)}={0}\)

\(\displaystyle-{4}{x}+{6}{y}+{10}{z}+{14}={0}\)