Prove that the line through $A(0, - 1, - 1)$ and $B(4,5,1)$ intersects the line through $C(3,9,4)$ and $D( - 4,4,4)$.

As we know, the cartesian equation of a line that passes through two points $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$

is
$\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}}$

Hence, the cartesian equation of line passes through $A(0, - 1, - 1)$

and $B(4,5,1)$ is $\frac{{x - 0}}{{4 - 0}} = \frac{{y + 1}}{{5 + 1}} = \frac{{z + 1}}{{1 + 1}}$

$\Rightarrow$ $\frac{x}{4} = \frac{{y + 1}}{6} = \frac{{z + 1}}{2}$

……..(i)
and cartesian equation of the line passes through $C(3,9,4)$ and $D( - 4,4,4)$ is $\frac{{x - 3}}{{ - 4 - 3}} = \frac{{y - 9}}{{4 - 9}} = \frac{{z - 4}}{{4 - 4}}$

$\Rightarrow$ $\frac{{x - 3}}{{ - 7}} = \frac{{y - 9}}{{ - 5}} = \frac{{z - 4}}{0}$

………(ii)
If the lines intersect, then shortest distance between both of them should be zero.

$\therefore$ Shortest distance between the lines

$= \frac{{\left| {\begin{array}{cccccccccccccccccccc}{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\end{array}} \right|}}{{\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}$

$= \frac{{\left| {\begin{array}{cccccccccccccccccccc}{3 - 0}&{9 + 1}&{4 + 1}\\4&6&2\\{ - 7}&{ - 5}&0\end{array}} \right|}}{{\sqrt {{{(6 \cdot 0 + 10)}^2} + {{( - 14 - 0)}^2} + {{( - 20 + 42)}^2}} }}$

$= \frac{{\left| {\begin{array}{cccccccccccccccccccc}3&{10}&5\\4&6&2\\{ - 7}&{ - 5}&0\end{array}} \right|}}{{\sqrt {100 + 196 + 484} }}$

$= \frac{{3(0 + 10) - 10(14) + 5( - 20 + 42)}}{{\sqrt {780} }}$

$= \frac{{30 - 140 + 110}}{{\sqrt {780} }} = 0$

So, the given lines intersect.