<s>[INST] Shortest distance between z axis and the line $x+y+2z=3, 2x+3y+4z+4=0$ z axis is : $x=0=y$ 
$x + B(y) = 0$          -->            (1)
$x+y+2z-3+ A (2x+3y+4z+4 ) = 0$              -->  (2)
These are two planes constructed through the two lines. We find the parallel planes so that it is easier to calculate the distance after that.
Therefore, for planes to be parallel :
$\frac{1+2A}{1} $  = $\frac{1+3A}{B}$ = $\frac{2+4A}{0}$
But now no values of A,B will satisfy this. 
How to proceed further ?
 [/INST] Solving the given equations for $x$ and $y$ in function of $z$ gives $x=13-2z$, $y=-10$. It follows that $$\sqrt{x^2+y^2}=\sqrt{(13-2z)^2+100}$$ is minimal when $z={13\over2}$, and the minimal value is $10$.
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