If these lines intersect we should be able to find x, y, z to satisfy all equations.
if x-4/5=2, then x=2 4/5=14/5; and y-1/2=2, making y=5/2.
x-1/2=14/5-1/2=28/10-5/10=23/10; y-2/3=5/2-2/3=15/6-4/6=11/6. So x-1/2 does not equal y-2/3. There is a lack of consistency, suggesting an error in the question, or misinterpretation of the question, or both. The lines cannot intersect. Perhaps the second set should be x-4/5=y-1/2=z. Let's see if changing 2 to z gives us a solution. It doesn't, because y-2/3=z-3/4=y-1/2-3/4=y-5/4 and 2/3 does not equal 5/4. Take x-1/2=y-2/3, then x-y=1/2-2/3=-1/6; also x-4/5=y-1/2 means x-y=4/5-1/2=3/10 and -1/6 does not equal 3/10, so there are other errors or misinterpretations.
Next, let's look at (x-1)/2=(y-2)/3=(z-3)/4 and (x-4)/5=(y-1)/2=2. This is another interpretation of the question.
So we have 3x-3=2y-4 and 3x-2y=-1; 2x-8=5y-5 and 2x-5y=3. Take the two equations: 3x-2y=-1 and 2x-5y=3. Doubling the first and tripling the second: 6x-4y=-2 and 6x-15y=9. Subtracting: 11y=-11 so y=-1 and 3x=2y-1=-3, making x=-1. Finally we use (x-1)/2=(z-3)/4 to find z. -1=(z-3)/4 so z=-1. Therefore x=y=z=-1. But (x-4)/5=(y-1)/2=? Let's see: -5/5=-2/2=-1=z. So the question should have read:
(x-1)/2=(y-2)/3=(z-3)/4 and (x-4)/5=(y-1)/2=z and the lines intersect at (-1,-1,-1).