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

We have the lines,

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

$\frac{x-4}{5}=\frac{y-1}{2}=z$

Let us denote these lines as L1and L2, such that

$L_{1}:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$

$L_{2}=\frac{x-4}{5}=\frac{y-1}{2}=z=\mu$

where $\lambda ,\mu \epsilon \mathbb{R}$

We must show that the lines L1and L2 intersect.

To show this, let us first find any point on line L1 and line L2

For L1:

$L_{1}:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$

$\Rightarrow \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$

$\Rightarrow \frac{x-1}{2}=\lambda,\frac{y-2}{3}=\lambda,\frac{z-3}{4}=\lambda$

We must find the values of x, y, and z. Therefore, let us take $\Rightarrow \frac{x-1}{2}=\lambda$

$\Rightarrow x-1=2\lambda$

$\Rightarrow x=2\lambda+1$

Take $\frac{y-2}{3}=\lambda$

$\Rightarrow y-2=3\lambda$

$\Rightarrow y=3 \lambda+2$

Take $\frac{z-3}{4}=\lambda$

$\Rightarrow z-3=4\lambda$

$\Rightarrow z=4\lambda+3...(i)$

Therefore, any point on L1 can be represented as $(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$.

Now,

For L2:

$L_{2}=\frac{x-4}{5}=\frac{y-1}{2}=z=\mu$

$\Rightarrow \frac{x-4}{5}=\frac{y-1}{2}=z=\mu$

$\Rightarrow \frac{x-4}{5}=\mu,\frac{y-1}{2}=\mu,z=\mu$

We must find the values of x, y, and z. Therefore,

Take $\frac{x-4}{5}=\mu$

$\Rightarrow x-4=5\mu$

$\Rightarrow x=5\mu+4$

Take $\frac{y-1}{2}=\mu$

$\Rightarrow y-1=2\mu$

$\Rightarrow y=2\mu+1$

Take $z=\mu$

$\Rightarrow z=\mu........(ii)$

Hence, any point on line L? can be represented as (5μ + 4, 2μ + 1, μ).

If lines L1 and L2 intersect, then there exist λ and μ such that

$\left ( 2\lambda+1,3\lambda+3,4\lambda+3 \right )\equiv \left ( 5\mu+4,2\mu+1,\mu \right )$

$\Rightarrow 2\lambda+1= 5\mu+4......(iii)$

$3\lambda+2=2\mu+1.....(iv)$

$4\lambda+3=\mu.....(iv)$

Substituting the value of μ from equation (v) into equation (iv),

$3\lambda+2=2\left ( 4\lambda+3 \right )+1$

$\Rightarrow 3\lambda+2=8\lambda+6+1$

$\Rightarrow 3\lambda+2=8\lambda+7$

$\Rightarrow 8\lambda-3\lambda=2-7$

$\Rightarrow 5\lambda=-5$

$\Rightarrow \lambda=-\frac{5}{5}$

$\Rightarrow \lambda=-1$

Putting this value of $\lambda$ in eq (v),

$4\left ( -1 \right )+3=\mu$

$\Rightarrow \mu=-4+3$

$\Rightarrow \mu=-1$

To check, we can substitute the values of $\lambda$ and $\mu$ in equation (iii), giving us:

$2(-1) + 1 = 5(-1) + 4$

$\Rightarrow -2 + 1 = -5 + 4$

$\Rightarrow -1 = -1$

Therefore $\lambda$ and $\mu$ also satisfy equation (iii).

So, the z-coordinate from equation (i),

$z=4\lambda +3$

$\Rightarrow z=4\left ( -1 \right )+3 \left [ \because \lambda=-1 \right ]$

$\Rightarrow z=-4+3$

$\Rightarrow z=-1$

And the z-coordinate from equation (ii),

$z=\mu$

$z=-1\left [ \because \mu=-1 \right ]$
So, the lines intersect at the point

$(5\mu + 4, 2\mu + 1, \mu) = (5(-1) + 4, 2(-1) + 1, -1).\\ Or, (5\mu + 4, 2\mu + 1, \mu) = (-5 + 4, -2 + 1, -1)\\ Or (5\mu + 4, 2\mu + 1, \mu) = (-1, -1, -1)$

Therefore the lines intersect at the point (-1, -1, -1).