Let two straight lines drawn from the origin $$\mathrm{O}$$ intersect the line $$3 x+4 y=12$$ at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ such that $$\triangle \mathrm{OPQ}$$ is an isosceles triangle and $$\angle \mathrm{POQ}=90^{\circ}$$. If $$l=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO}^2$$, then the greatest integer less than or equal to $$l$$ is :

The vertices of a triangle are $$\mathrm{A}(-1,3), \mathrm{B}(-2,2)$$ and $$\mathrm{C}(3,-1)$$. A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

Let $$A(a, b), B(3,4)$$ and $$C(-6,-8)$$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $$P(2 a+3,7 b+5)$$ from the line $$2 x+3 y-4=0$$ measured parallel to the line $$x-2 y-1=0$$ is

Let $$\alpha, \beta, \gamma, \delta \in \mathbb{Z}$$ and let $$A(\alpha, \beta), B(1,0), C(\gamma, \delta)$$ and $$D(1,2)$$ be the vertices of a parallelogram $$\mathrm{ABCD}$$. If $$A B=\sqrt{10}$$ and the points $$\mathrm{A}$$ and $$\mathrm{C}$$ lie on the line $$3 y=2 x+1$$, then $$2(\alpha+\beta+\gamma+\delta)$$ is equal to