If the equations $3 x^2+2 h x y-3 y^2=0$ and $3 x^2+2 h x y-3 y^2+2 x-4 y+c=0$ represent the four sides of a square, then $\frac{h}{c}=$

$(a, b)$ are the new coordinates of the point $(2,3)$ after shifting the origin to the point $(3,2)$ by translation of axes. If $(c, d)$ are the new coordinates of the point $(a, b)$ after rotating the axes through an angle $\frac{\pi}{4}$ about the origin in the anti-clockwise direction, then $d-c=$

The lines $x+y+4=0, x-2 y-4=0$ and $3 x+4 y-2=0$

The area of the triangle formed by the line $L$ with the coordinate axes is 12 sq. units. If $L$ passes through the point $(12,4)$ and the product $P$ of $X$ - intercept of $L$ and square of the $Y$-intercept of $L$ is negative, then $P=$