$L \equiv x \cos \alpha+y \sin \alpha-p=0$ represents a line perpendicular to the line $x+y+1=0$. If $p$ is positive, $\alpha$ lies in the fourth quadrant and perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line, $L=0$ is 5 units, then $p=$

$(-2,-1),(2,5)$ are two vertices of a triangle and $\left(2, \frac{5}{3}\right)$ is its orthocenter. If $(m, n)$ is the third vertex of that triangle, then $m+n$ is equal to.

$L_1 \equiv 2 x+y-3=0$ and $L_2 \equiv a x+b y+c=0$ are two equal sides of an isosceles triangle. If $L_3 \equiv x+2 y+1=0$ is the third side of this triangle and $(5,1)$ is a point on $L_2=$ 0 , then $\frac{b^2}{|a c|}=$

The slope of one of the pair of lines $2 x^2+h x y+6 y^2=0$ is thrice the slope of the other line, then $h=$