Show that the lines $\;$ $\dfrac{x-1}{1} = \dfrac{y + 1}{-1} = \dfrac{z}{3}$ $\;$ and $\;$ $\dfrac{x-2}{1} = \dfrac{y - 1}{2} = \dfrac{-z -1}{1}$ $\;$ intersect and find their point of intersection.
Equations of given lines are
$\dfrac{x-1}{1} = \dfrac{y + 1}{-1} = \dfrac{z}{3}$ $\;\;\; \cdots \; (1a)$
$\dfrac{x-2}{1} = \dfrac{y - 1}{2} = \dfrac{-z -1}{1}$ $\;\;\; \cdots \; (1b)$
Let $\ell_1$, $m_1$ and $n_1$ be the direction ratios of equation $(1a)$.
Let $\ell_2$, $m_2$ and $n_2$ be the direction ratios of equation $(1b)$.
Comparing equations $(1a)$ and $(1b)$ with
$\dfrac{x - x_1}{\ell_1} = \dfrac{y - y_1}{m_1} = \dfrac{z - z_1}{n_1}$ $\;$ and $\;$ $\dfrac{x - x_2}{\ell_2} = \dfrac{y - y_2}{m_2} = \dfrac{z - z_2}{n_2}$ $\;$ respectively, we have
$x_1 = 1$, $y_1 = -1$, $z_1 = 0$; $\;\;$ $x_2 = 2$, $y_2 = 1$, $z_2 = -1$
$\ell_1 = 1$, $m_1 = -1$, $n_1 = 3$; $\;\;$ $\ell_2 = 1$, $m_2 = 2$, $n_2 = -1$
The condition for intersecting lines is
$\begin{vmatrix}
x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\
\ell_1 & m_1 & n_1 \\
\ell_2 & m_2 & n_2
\end{vmatrix} = 0$ $\;\;\; \cdots \; (2)$
Now,
$\begin{aligned}
\begin{vmatrix}
x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\
\ell_1 & m_1 & n_1 \\
\ell_2 & m_2 & n_2
\end{vmatrix} & = \begin{vmatrix}
2 - 1 & 1 + 1 & -1 - 0 \\
1 & -1 & 3 \\
1 & 2 & -1
\end{vmatrix} \\\\
& = \begin{vmatrix}
1 & 2 & -1 \\
1 & -1 & 3 \\
1 & 2 & -1
\end{vmatrix} \\\\
& = 1 \left(1 - 6\right) - 2 \left(-1 -3\right) -1 \left(2 + 1\right) \\\\
& = -5 + 8 -3 = 0
\end{aligned}$
$\because$ $\;$ equation $(2)$ is satisfied by the given lines, the given lines are intersecting.
Let $\;$ $\dfrac{x-1}{1} = \dfrac{y + 1}{-1} = \dfrac{z}{3} = \lambda$ $\;\;\; \cdots \; (3a)$
$\therefore$ $\;$ Any point on line $(3a)$ is of the form $\;$ $\left(\lambda + 1, \; - \lambda - 1, \; 3 \lambda\right)$
Let $\dfrac{x-2}{1} = \dfrac{y - 1}{2} = \dfrac{-z -1}{1} = \mu$ $\;\;\; \cdots \; (3b)$
$\therefore$ $\;$ Any point on line $(3b)$ is of the form $\;$ $\left(\mu + 2, \; 2\mu + 1, \; -\mu - 1\right)$
$\because$ $\;$ The two given lines are intersecting, for some $\lambda$, $\mu$
$\left(\lambda + 1, \; - \lambda - 1, \; 3 \lambda\right) = \left(\mu + 2, \; 2\mu + 1, \; -\mu - 1\right)$
$\implies$ $\lambda + 1 = \mu + 2$ $\;$ i.e. $\;$ $\lambda - \mu = 1$ $\;\;\; \cdots \; (4a)$
and $\;$ $3 \lambda = -\mu - 1$ $\;$ i.e. $\;$ $3 \lambda + \mu = -1$ $\;\;\; \cdots \; (4b)$
From equations $(4a)$ and $(4b)$, $\;$ $\lambda = 0$, $\;$ $\mu = -1$
Taking either $\lambda = 0$ or $\mu = -1$, the point of intersection of the two lines is $\;$ $\left(1, -1, 0\right)$