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

(A) The Cartesian equation of a line passing through two points $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ is given by $\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}$.
The equation of the line passing through $A(0,-1,-1)$ and $B(4,5,1)$ is $\frac{x-0}{4-0} = \frac{y+1}{5+1} = \frac{z+1}{1+1}$,which simplifies to $\frac{x}{4} = \frac{y+1}{6} = \frac{z+1}{2} = \lambda$ (say). Any point on this line is $(4\lambda, 6\lambda-1, 2\lambda-1)$.
The equation of the line passing through $C(3,9,4)$ and $D(-4,4,4)$ is $\frac{x-3}{-4-3} = \frac{y-9}{4-9} = \frac{z-4}{4-4}$,which simplifies to $\frac{x-3}{-7} = \frac{y-9}{-5} = \frac{z-4}{0} = \mu$ (say). Any point on this line is $(-7\mu+3, -5\mu+9, 4)$.
For the lines to intersect,there must exist $\lambda$ and $\mu$ such that the coordinates are equal:
$2\lambda-1 = 4 \Rightarrow 2\lambda = 5 \Rightarrow \lambda = 2.5$.
Substituting $\lambda = 2.5$ into the first two coordinates: $x = 4(2.5) = 10$,$y = 6(2.5)-1 = 14$.
Now,for the second line: $-7\mu+3 = 10 \Rightarrow -7\mu = 7 \Rightarrow \mu = -1$.
Checking $y$: $-5(-1)+9 = 5+9 = 14$. Since the coordinates match,the lines intersect.