Find the value of p, if the straight lines 3x + 7y – 1= 0 and 7x – py + 3 = 0 are mutually perpendicular.

$\text{ If }{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\text{ and }{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0\text{ are }$mutually perpendicular then ${\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2=0$
$$\begin{gathered} \text{ Given lines }3\mathrm{x}+7\mathrm{y}-1=0 ...\left(1\right) \\ \text{ And }7\mathrm{x}-\mathrm{py}+3=0 ...\left(2\right) \end{gathered}$$
Since (1) and (2) are mutually perpendicular then
$\left(3\right)\cdot \left(7\right)+7(-\mathrm{p})=0\Rightarrow 21-7\mathrm{p}=0\Rightarrow \mathrm{p}=3$