The shortest distance between the z - axis and the line x + y + 2 z - 3 = 0 = 2 x + 3 y + 4 z - 4 , is

We have,

$x+y+2z-3=0=2x+3y+4z-4$

Change it into symmetrical form

Put $z=1,$ then

$x+y-1=0$

$2x+3y=0$

On solving, we get

$x=3, y=-2, z=1$

$\left(3, -2 , 1\right)$

Direction ratio of the line is

$=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 1& 1& 2\\ 2& 3& 4\end{array}\right|=-2\widehat{i}+\widehat{k}$

Hence, equation of line in symmetrical form is

$\frac{x-3}{-2}= \frac{y+2}{0}=\frac{z-1}{1}$

Equation of  $z$ axis is, $\frac{x}{0}= \frac{y}{0}= \frac{z}{1}$

Shortest Distance $=\frac{\left|\begin{array}{ccc}3& -2& 1\\ -2& 0& 1\\ 0& 0& 1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 0& 1\\ 0& 0& 1\end{array}\right|}$

$=\left|\frac{-4}{2\hat{j}}\right|=2$