(i) Let the coordinate of point $P\left(x_1, y_1\right)$ on the line $x+5 y=13 i$.e.,

$$P\left(13-5 y_1, y_1\right)$$

$\therefore$ Distance of $P$ from the line $12 x-5 y+26=0$,

$$\begin{array}{l} & 2 =\left|\frac{12\left(13-5 y_1\right)-5 y_1+26}{\sqrt{144+25}}\right| \\ \Rightarrow \quad & 2 = \pm \frac{156-60 y_1-5 y_1+26}{13} \\ \Rightarrow \quad & -65 y_1 =-156 \quad \text{[taking positive sign]}\\ \Rightarrow \quad& y_1 =\frac{156}{65}=\frac{12}{5} \\ \because \quad & x_1=13-5 y_1 \\ & =13-12=1 \end{array}$$

So, the coordinate of is $P\left(1, \frac{12}{5}\right)$.

Similarly, the coordinates of $Q$ are $\left(-3, \frac{16}{5}\right)\quad \text{[taking negative sign]}$.

(ii) Let coordinates of the point on the line $x+y=4$ be $\left(4-y_1, y_1\right)$.

Distance from the line $4 x+3 y-10=0$.

$$\begin{aligned} & 1=\left|\frac{4\left(4-y_1\right)+3 y_1-10}{\sqrt{16+9}}\right| \\ \Rightarrow & 1= \pm \frac{16-4 y_1+3 y_1-10}{5} \quad\text{[taking negative sign]}\\ \Rightarrow & 5=6-y_1 \\ \Rightarrow & y_1=1 \end{aligned}$$

If $y_1=1$, then $x_1=3$

So, the point is $(3,1)$.

Similarly, taking negative sign the point is $(-7,11)$.

(iii) Given point $A(-2,5)$ and $B(3,1)$.

Now, the point $P$ divides line joining the point $A$ and $B$ in $1: 2$.

$$\begin{array}{ll} \because & x_1=\frac{1 \cdot 3+2(-2)}{1+2}=\frac{3-4}{3}=\frac{-1}{3} \\ \text { and } & y_1=\frac{1 \cdot 1+2 \cdot 5}{1+2}=\frac{11}{3} \end{array}$$

So, the coordinates of $P$ are $\left(\frac{-1}{3}, \frac{11}{3}\right)$.

Thus, the point $Q$ divided the line joining $A$ to $B$ in $2: 1$.

$$\begin{array}{ll} \because & x_2=\frac{2 \cdot 3+1(-2)}{2+1}=\frac{4}{3} \\ \text { and } & y_2=\frac{2 \cdot 1+1 \cdot 5}{2+1}=\frac{7}{3} \end{array}$$

Hence, the coordinates of $Q$ are $\left(\frac{4}{3}, \frac{7}{3}\right)$.

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

(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).