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**Hint:**Equation in this form \[\dfrac{x-a}{l}=\dfrac{y-b}{m}=\dfrac{z-c}{n}\] is a line passing through point (a, b, c) with direction cosines (l, m, n). Then the equation of the plane passing through a line will be of the form a’(x-a) + b’(y-b) + c’(z-c) = 0 where a’, b’ and c’ are the direction cosines of the normal line of the plane.

**Complete step-by-step solution:**

From the question, the given two lines are \[\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z}{3}\] and \[\dfrac{x}{2}=\dfrac{y-2}{-1}=\dfrac{z+1}{3}\].

Here, we can see that the given two lines have same direction cosines which implies that they are parallel to each other.

Now, let’s find the equation of the plane.

Equation of any plane containing the first line is given by

p (x – 1) + q (y + 1) + r(z) = 0 -------------(1)

where 2p – q + 3r = 0 -----------------(2)

We know that the plane passes through the second line if (0, 2, -1), which is the point on the second line, lies on it.

So, on substituting the point (0, 2, -1) in the equation (1), we get

-p + 3q – r = 0 -------------(3)

Then on solving equations (2) and (3), we get

\[\dfrac{p}{-8}=\dfrac{q}{-1}=\dfrac{r}{5}\] or \[\dfrac{p}{8}=\dfrac{q}{1}=\dfrac{r}{-5}\]

Hence, the required equation of plane is

8 (x – 1) + (y + 1) – 5z = 0

\[\Rightarrow \] 8x + y – 5z – 7 = 0.

**Note:**While solving such types of questions, it is necessary to check whether they are parallel to each other or perpendicular to each other. If the two lines are perpendicular then we cannot form a plane from these two lines. If the two lines are non-parallel coplanar like \[\dfrac{x-x_1}{b_1}=\dfrac{y-y_1}{b_2}=\dfrac{z-z_1}{b_3}\] and \[\dfrac{x-x_2}{d_1}=\dfrac{y-y_2}{d_2}=\dfrac{z-z_2}{d_3}\], then the equation of plane is given by

\[\left| \begin{matrix}

x-x_1 & y-y_1 & z-z_1 \\

b_1 & b_2 & b_3 \\

d_1 & d_2 & d_3 \\

\end{matrix} \right|=0\] or \[\left| \begin{matrix}

x-x_2 & y-y_2 & z-z_2 \\

b_1 & b_2 & b_3 \\

d_1 & d_2 & d_3 \\

\end{matrix} \right|=0\] .

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