Show that the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}$ intersect. Also,find their point of intersection.

(A) Let the equations of the lines be:
$L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} = \lambda$
$L_2: \frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1} = \mu$
Any point on $L_1$ is $(2\lambda+1, 3\lambda+2, 4\lambda+3)$ and any point on $L_2$ is $(5\mu+4, 2\mu+1, \mu)$.
If the lines intersect,there exist $\lambda$ and $\mu$ such that:
$2\lambda+1 = 5\mu+4 \Rightarrow 2\lambda - 5\mu = 3$ $(1)$
$3\lambda+2 = 2\mu+1 \Rightarrow 3\lambda - 2\mu = -1$ $(2)$
$4\lambda+3 = \mu$ $(3)$
Substitute $(3)$ into $(1)$:
$2\lambda - 5(4\lambda+3) = 3$
$2\lambda - 20\lambda - 15 = 3$
$-18\lambda = 18 \Rightarrow \lambda = -1$
Using $\lambda = -1$ in $(3)$:
$\mu = 4(-1)+3 = -1$
Check these values in $(2)$:
$3(-1) - 2(-1) = -3 + 2 = -1$. This satisfies equation $(2)$.
Since the values satisfy all equations,the lines intersect.
Point of intersection using $\lambda = -1$:
$x = 2(-1)+1 = -1$
$y = 3(-1)+2 = -1$
$z = 4(-1)+3 = -1$
Thus,the point of intersection is $(-1, -1, -1)$.