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

If shortest distance between the lines is zero, then they intersect.

Let us assume ${x_1} = 1,{y_1} = 2,{z_1} = 3$ and ${a_1} = 2,{b_1} = 3,{c_1} = 4$

Also, ${x_2} = 4,{y_2} = 1,{z_2} = 0$

and ${a_2} = 5,{b_2} = 2,{c_2} = 1$

If two lines intersect, then shortest distance between them should be zero.
$\therefore$

Shortest distance between two given lines

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

$= \frac{{\left| {\begin{array}{cccccccccccccccccccc}{4 - 1}&{1 - 2}&{0 - 3}\\2&3&4\\5&2&1\end{array}} \right|}}{{\sqrt {{{(3 \cdot 1 - 2 \cdot 4)}^2} + {{(4 \cdot 5 - 1 \cdot 2)}^2} + {{(2 \cdot 2 - 5 \cdot 3)}^2}} }}$

$= \frac{{\left| {\begin{array}{cccccccccccccccccccc}3&{ - 1}&{ - 3}\\2&3&4\\5&2&1\end{array}} \right|}}{{\sqrt {25 + 324 + 121} }}$
$= \frac{{3(3 - 8) + 1(2 - 20) - 3(4 - 15)}}{{\sqrt {470} }}$

$= \frac{{ - 15 - 18 + 33}}{{\sqrt {470} }} = \frac{0}{{\sqrt {470} }} = 0$

Therefore, the given two lines are intersecting.

For finding their point of intersection for first line,

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

$\Rightarrow$ $x = 2\lambda + 1,y = 3\lambda + 2$ and $z = 4\lambda + 3$

Since, the lines are intersecting. So, let us put these values in the equation of another line.

Thus, $\frac{{2\lambda + 1 - 4}}{5} = \frac{{3\lambda + 2 - 1}}{2} = \frac{{4\lambda + 3}}{1}$

$\Rightarrow$ $\frac{{2\lambda - 3}}{5} = \frac{{3\lambda + 1}}{2} = \frac{{4\lambda + 3}}{1}$

$\Rightarrow$ $\frac{{2\lambda - 3}}{5} = \frac{{4\lambda + 3}}{1}$
$\Rightarrow$ $2\lambda - 3 = 20\lambda + 15$

$\Rightarrow$ $18\lambda = - 18 = - 1$

So, the required point of intersection is
$x = 2( - 1) + 1 = - 1$
$y = 3( - 1) + 2 = - 1$
$z = 4( - 1) + 3 = - 1$

Thus, the lines intersect at (-1,-1,-1)