Find the values of p in (i) to (iv) and

(i) Given pair of linear equations is

$3 x-y-5=0$ $\ldots$ (i)

and $\quad 6 x-2 y-p=0$ ....(ii)

On comparing with $a x+b y+c=0$, we get

$a_{1}=3, b_{1}=-1$

and $c_{1}=-5$ [from Eq. (i)]

$a_{2}=6, b_{2}=-2$

and $c_{2}=-p$ [from Eq. (ii)]

Since, the lines represented by these equations are parallel, then

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\frac{3}{6}=\frac{-1}{-2} \neq \frac{-5}{-p}$

Taking last two parts, we get $\frac{-1}{-2} \neq \frac{-5}{-p}$

$\Rightarrow \quad \frac{1}{2} \neq \frac{5}{p}$

$\Rightarrow \quad p \neq 10$

Hence, the given pair of linear equations are parallel for all real values of $p$ except 10 i.e., $p \in R-\{10\}$.

(ii) Given pair of linear equations is

$-x+p y-1=0$ $\ldots$ (i)

and $p x-y-1=0$ $\ldots$ (ii)

On comparing with $a x+b y+c=0$, we get

$a_{1}=-1, b_{1}=p$ and $c_{1}=-1 \quad$ [from Eq. (i)]

$a_{2}=p_{1} b_{2}=-1$ and $c_{2}=-1$ [from Eq. (ii)]

Since, the pair of linear equations has no solution i.e., both lines are parallel to each other.

$\therefore \quad \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \frac{-1}{p}=\frac{p}{-1} \neq \frac{-1}{-1}$

Taking last two parts, we get

$\frac{p}{-1} \neq \frac{-1}{-1}$

$\Rightarrow \quad p \neq-1$

Taking first two parts, we get

$\frac{-1}{p}=\frac{p}{-1}$

$\Rightarrow \quad p^{2}=1$

$\Rightarrow \quad p=\pm 1$

but $p \neq-1$

$\therefore \quad p=1$

Hence, the given pair of linear equations has no solution for $p=1$.

(III) Given, pair of linear equations is

$-3 x+5 y-7=0$ $\ldots$ (i)

and $\quad 2 p x-3 y-1=0 \quad$...(ii)

On comparing with $a x+b y+c=0$, we get

$a_{1}=-3, b_{1}=5$

$\begin{array}{lll}\text { and } & c_{1}=-7 & \text { [from Eq. (i)] }\end{array}$

$a_{2}=2 p_{1} b_{2}=-3$

$\begin{array}{lll}\text { and } & c_{2}=-1 & \text { [from Eq. (ii)] }\end{array}$

Since, the lines are intersecting at a unique point i.e., it has a unique solution.

$\therefore$ $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

$\Rightarrow \quad \frac{-3}{2 p} \neq \frac{5}{-3}$

$\Rightarrow \quad 9 \neq 10 p$

$\Rightarrow \quad p \neq \frac{9}{10}$

Hence, the lines represented by these equations are intersecting at a unique point for all real values of $p$ except $\frac{9}{10}$

(Iv) Given pair of linear equations is

$2 x+3 y-5=0$

and $\quad p x-6 y-8=0 \quad$... (ii)

On comparing with $a x+b y+c=0$, we get

$a_{1}=2, b_{1}=3$

and $\quad c_{1}=-5 \quad$ [from Eq. (i)]

$a_{2}=p_{1} \quad b_{2}=-6$

$\begin{array}{lcc}\text { and } & c_{2}=-8 & \text { [from Eq. (ii)] }\end{array}$

Since, the pair of linear equations has a unique solution.

$\therefore$ $\frac{a_{1}}{a_{2}} \neq \frac{D_{1}}{b_{2}}$

$\Rightarrow$ $\frac{2}{p} \neq \frac{3}{-6}$

$\Rightarrow$ $p \neq-4$

Hence, the pair of linear equations has a unique solution for all values of $p$ except $-4$ i.e.,

$p \in R-\{-4,$,

(v) Given pair of linear equations is

$2 x+3 y=7$ $\ldots(1)$

and $\quad 2 p x+p y=28-q y$

$\Rightarrow \quad 2 p x+(p+q) y=28 \quad$...(ii)

On comparing with $a x+b y+c=0$, we get

$a_{1}=2, b_{1}=3$

and $c_{1}=-7 \quad$ [from Eq. (i)]

$a_{2}=2 p_{1} b_{2}=(p+q)$

and   $c_{2}=-28$ [from Eq. (ii)] 

Since, the pair of equations has infinitely many solutions $i$ i. $e$., both lines are coincident.

$\therefore \quad \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \quad \frac{2}{2 p}=\frac{3}{(p+q)}=\frac{-7}{-28}$

Taking first and third parts, we get

$\frac{2}{2 p}=\frac{-7}{-28}$

$\Rightarrow \quad \frac{1}{p}=\frac{1}{4}$

$\Rightarrow \quad p=4$

Again, taking last two parts, we get 

$\frac{3}{p+q}=\frac{-7}{-28} \Rightarrow \frac{3}{p+q}=\frac{1}{4}$

$\Rightarrow \quad p+q=12$

$\Rightarrow \quad 4+q=12 \quad[\because p=4]$

$\therefore \quad q=8$

Here, we see that the values of $p=4$ and $q=8$ satisfies all three parts.

Hence, the pair of equations has infinitely many solutions for the values of $p=4$ and $q=8$