The equation of line which equally inclined to the axis and passes through common points of family of lines $4acx+y(ab+bc+ca-abc)+abc=0$  where a, b and $c>0$ are in  $H.P$ is 

Given line is $4acx+y(ab+bc+ca-abc)+abc=0$  

dividing both sides by abc 

$$\begin{gathered} \frac{4}{b}x+y(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-1)+1=0 \\ Given a,b,c are in H.P \end{gathered}$$

Line can be written as  

$$\begin{gathered} \frac{4}{b}x+\frac{3}{b}y+1-y=0 \\ Equatioin is 4x+(3-b)y+b=0 \\ \sin ce line is equally inclined to the axes then slope of line is 1 or -1 \\ \Rightarrow b=7 or -1 \\ Required line is y-x=\frac{7}{4} or y+x=\frac{1}{4} \end{gathered}$$
Hence, the required line is $y-x=\frac{7}{4}$ 
Hence, the correct answer is (A).