Question

The shortest distance between the line $x+y+2z-3=2x+3y+4z-4=0$ and $xz$-plane is units

Answer 1

$x+y+2z=3\cdots \left(1\right)2x+3y+4z=4\Rightarrow x+\frac{3}{2}y+2z=2\cdots \left(2\right)$
Using equation $\left(1\right)$ and $\left(2\right)$, we get
$y=-2$

Hence, the shortest distance will be $2$ units.