The given two lines are,

\(v_1=(0,1,-2)+t(2,3,-1)\) and

\(v_2=(2,-1,0)+t(2,3,-1)\)

Let A=(0,1,-2)

B=(2,-1,0)

Thus the plane containing these two lines will contain A and B and hence the vector AB=(2,-2,2)

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

Thus \(\vec{n}=\vec{AB}\times\text{direction ratio of the line}\)

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

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

Thus the equation of the plane can be written as,

\([(x-0)i+(y-1)j+(z+2)\cdot\vec{n}=0\)

\(-4(x-0)+6(y-1)+10(z+2)=0\)

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

\(v_1=(0,1,-2)+t(2,3,-1)\) and

\(v_2=(2,-1,0)+t(2,3,-1)\)

Let A=(0,1,-2)

B=(2,-1,0)

Thus the plane containing these two lines will contain A and B and hence the vector AB=(2,-2,2)

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

Thus \(\vec{n}=\vec{AB}\times\text{direction ratio of the line}\)

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

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

Thus the equation of the plane can be written as,

\([(x-0)i+(y-1)j+(z+2)\cdot\vec{n}=0\)

\(-4(x-0)+6(y-1)+10(z+2)=0\)

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