If lines $\frac{x-3}{2}=\frac{y+1}{-3}=\frac{z+a}{P}$  and $\frac{x+2}{2}=\frac{y-4}{4}=\frac{z+5}{2}$  are perpendicular coplanar lines, then value of a + p is

$\left|\begin{array}{ccc}5& -5& 5-a\\ 2& -3& p\\ 2& 4& 2\end{array}\right|=0$

$\Rightarrow 60-30p-14a=0$      ………. (1)

Lines are perpendicular $\therefore 4-12+2p=0$   ……….. (2)

From (1) & (2) $P=4,a=\frac{-30}{7}$