If the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent. Then find ‘p’.

$$\begin{gathered} Given lines are \\ x+p=0,y+2=0 and 3x+2y+5=0 \end{gathered}$$

$\text{ Let }\mathrm{p}(\mathrm{\alpha },\mathrm{\beta })\text{ be point of concurrence }$then
$\begin{array}{l}\mathrm{\alpha }+\mathrm{p}=0\dots \left(1\right);\mathrm{\beta }+2=0\dots .\left(2\right)\\ \Rightarrow \mathrm{\alpha }=-\mathrm{p},\mathrm{\beta }=-2\\ \sin ce given lines are concurrent then\\ 3\mathrm{\alpha }+2\mathrm{\beta }+5=0\dots .\left(3\right)\Rightarrow 3p+2(-2)+5=0\\ \Rightarrow 3p+1=0\\ \Rightarrow p=\frac{-1}{3}\end{array}$