Brilliant Family

Let $L(x,y) \equiv x-y+3=0$ . We have,

$(3\sec \theta + 5\csc \theta)x+(7\sec \theta-3\csc \theta)y+11(\sec \theta - \csc \theta)=0$

$\implies \sec(\theta) (3x + 7y +11 ) + \csc(\theta) (5x-3y-11) =0 \quad \ (1)$

Family of lines $(1)$ will always pass through the intersection of the lines $3x + 7y +11 =0$ and $5x-3y-11 = 0$ which is $(1,-2)$

Let $B \equiv (1,-2)$ and $A \equiv (3,4)$

Now, $m_{AB} = 3 \neq m_{L}$ , so $AB$ and $L$ are not parallel. Further,

$L(3,4) \cdot L(1,-2) > 0 \implies \ \mathrm{Points \ lie \ on \ the \ same \ side \ of \ the \ line \ L}$

Thus there must be a point on the extended line segment $AB$ intersecting the line $L$ . Let this point be $C$ .

$\implies |PA - PB| \leq |AB| = 2\sqrt{10} \ (\mathrm{Triangular \ Inequality})$

where the equality holds at point $C$ .

$\therefore n = \boxed{10}$