Question:

### Pair (i)
Only one equation \(5x-4y+8=0\) is given. To apply the ratio test we need two equations, therefore the nature of the pair cannot be decided with the information provided.

### Pair (ii)
We have two distinct equations:
\[
\begin{aligned}
&\text{(a)}\;9x+3y+12=0 \quad\Rightarrow\; a_1=9,\; b_1=3,\; c_1=12,\\[2mm]
&\text{(b)}\;7x+6y-9=0 \quad\Rightarrow\; a_2=7,\; b_2=6,\; c_2=-9.
\end{aligned}
\]
Step 1 – Form the ratios
\[
\frac{a_1}{a_2}=\frac{9}{7},\qquad \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2},\qquad \frac{c_1}{c_2}=\frac{12}{-9}= -\frac{4}{3}.
\]
Step 2 – Compare
Since \(\frac{9}{7}
eq\frac{1}{2}
eq-\frac{4}{3}\), the three ratios are all different. Hence the two lines intersect at a single point.

Additional observation
The third equation \(18x+6y+24=0\) is exactly twice the first equation (multiply \(9x+3y+12=0\) by 2). Therefore the line represented by \(18x+6y+24=0\) is coincident with the line \(9x+3y+12=0\).

### Pair (iii)
Equations:
\[
\begin{aligned}
&\text{(a)}\;6x-3y+10=0 \quad\Rightarrow\; a_1=6,\; b_1=-3,\; c_1=10,\\[2mm]
&\text{(b)}\;2x-y+9=0 \quad\Rightarrow\; a_2=2,\; b_2=-1,\; c_2=9.
\end{aligned}
\]
Step 1 – Form the ratios
\[
\frac{a_1}{a_2}=\frac{6}{2}=3,\qquad \frac{b_1}{b_2}=\frac{-3}{-1}=3,\qquad \frac{c_1}{c_2}=\frac{10}{9}
eq3.
\]
Step 2 – Compare
Here \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=3\) but \(\frac{c_1}{c_2}
eq3\). Hence the two lines are parallel (they have the same slope but different intercepts) and are not coincident.

Conclusion
- Pair (i): Insufficient data – cannot decide.
- Pair (ii): The lines \(9x+3y+12=0\) and \(7x+6y-9=0\) intersect at a point. The line \(18x+6y+24=0\) is coincident with \(9x+3y+12=0\).
- Pair (iii): The lines \(6x-3y+10=0\) and \(2x-y+9=0\) are parallel (distinct).