(i) Given equation of the line is

$$(2 x+3 y+4)+\lambda(6 x-y+12)=0\quad \text{.... (i)}$$

If line is parallel to $Y$-axis i.e., it is perpendicular to $X$-axis

$$\therefore \quad \text { Slope }=m=\tan 90 \Upsilon=\infty$$

From line (i), $x(2+6 \lambda)+y(3-\lambda)+4+12 \lambda=0$

and slope $$=\frac{-(2+6 \lambda)}{3-\lambda}$$

$$\begin{array}{ll} \Rightarrow & \frac{-2-6 \lambda}{3-\lambda}=\infty \\ \Rightarrow & \frac{-2-6 \lambda}{3-\lambda}=\frac{1}{0} \Rightarrow \lambda=3 \end{array}$$

(ii) If the line (i) is perpendicular to the line $7 x+y-4=0$ or $y=-7 x+4$

$$\begin{aligned} \because \quad & \frac{-(2+6 \lambda)}{(3-\lambda)}(-7) =-1 \\ \Rightarrow \quad & 14+42 \lambda =-3+\lambda \\ \Rightarrow \quad & 41 \lambda =-17 \\ \Rightarrow \quad & \lambda =-\frac{17}{41} \end{aligned}$$

(iii) If the line (i) passes through the point $(1,2)$.

$$\begin{aligned} \text{Then, }\quad (2+6+4)+\lambda(6-2+12) & =0 \\ 12+16 \lambda=0 \Rightarrow \lambda & =-\frac{3}{4} \end{aligned}$$

(iv) If the line is parallel to $X$-axis the slope $=0$.

Then, $$\frac{-(2+6 \lambda)}{3-\lambda}=0$$

$$\Rightarrow \quad-(2+6 \lambda)=0 \Rightarrow \lambda=-\frac{1}{3}$$

So, the correct matches are (i) $\rightarrow$ (d), (ii) $\rightarrow$ (c), (iii) $\rightarrow$ (a), (iv) $\rightarrow$ (b).

$$\begin{aligned} \text{Given equation of the lines are}\quad & 2 x-3 y=0 \quad \text{... (i)}\\ \text{and}\quad & 4 x-5 y=2\quad \text{... (ii)} \end{aligned}$$

From Eq. (i), put $x=\frac{3 y}{2}$ in Eq. (ii), we get

$$\begin{aligned} 4 \cdot \frac{(3 y)}{2}-5 y & =2 \\ \Rightarrow \quad 6 y-5 y & =2 \\ \Rightarrow \quad y & =2 \end{aligned}$$

Now, put $y=2$ in Eq. (i), we get $$x=3$$

So, the intersection points are $(3,2)$.

(i) The equation of the line passes through the point $(3,2)$ and $(2,1)$, is

$$\begin{array}{rlrl} & y-2 & =\frac{1-2}{2-3}(x-3) \\ \Rightarrow & y-2 & =(x-3) \\ \Rightarrow & x-y-1 & =0 \end{array}$$

(ii) If the required line is perpendicular to the line $x+2 y+1=0$

$\because$ Slope of the required line $=2$

$\therefore$ Equation of the line is

$$y-2=2(x-3)$$

$$\Rightarrow \quad 2 x-y-4=0$$

(iii) If the required line is parallel to the line $3 x-4 y+5=0$, then the slope of the required line $=\frac{3}{4}$

$\therefore$ Equation of the required line is

$$y-2=\frac{3}{4}(x-3)$$

$$\begin{array}{l} \Rightarrow & 4 y-8=3 x-9 \\ \Rightarrow & 3 x-4 y-1=0 \end{array}$$

(iv) If the line is equally inclined to the $X$-axis, then

$$m= \pm \tan 45 \Upsilon= \pm 1$$

$\therefore$ Equation of the line is

$$\begin{array}{l} & y-2 & =-1(x-3) \text{[taking negative value]}\\ \Rightarrow & y-2 & =-x+3 \\ \Rightarrow & x+y-5 & =0 \end{array}$$

So, the correct matches are (a) $\rightarrow$ (iii), (b) $\rightarrow$ (i), (c) $\rightarrow$ (iv), (d) $\rightarrow$ (ii).