The shortest distance between the lines $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$ and $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$ is

Shortest distance between two lines

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given as

$\frac{\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{\left(a_1{b}_2-a_2{b}_1\right)}^2+{\left(b_1{c}_2-b_2{c}_1\right)}^2+{\left(c_1{a}_2-c_2{a}_1\right)}^2}}$

$=\frac{\left|\begin{array}{ccc}5-\left(-3\right)& 2-\left(-5\right) & 4-1\\ 1& 2& -3\\ 1& 4& -5\end{array}\right|}{\sqrt{{\left(4-2\right)}^2+{\left(-10+12\right)}^2+{\left(-3+5\right)}^2}}$

$=\frac{\left|\begin{array}{ccc}8& 7& 3\\ 1& 2& -3\\ 1& 4& -5\end{array}\right|}{\sqrt{{\left(2\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2}}$

$=\frac{\left|8\left(-10+12\right)-7\left(-5+3\right)+3\left(4-2\right)\right|}{\sqrt{4+4+4}}$

$=\frac{16+14+6}{\sqrt{12}}=\frac{36}{2\sqrt{3}}$

$=\frac{18}{\sqrt{3}}=6\sqrt{3}$