Let $$\mathrm{A}(-1,1)$$ and $$\mathrm{B}(2,3)$$ be two points and $$\mathrm{P}$$ be a variable point above the line $$\mathrm{AB}$$ such that the area of $$\triangle \mathrm{PAB}$$ is 10. If the locus of $$\mathrm{P}$$ is $$\mathrm{a} x+\mathrm{by}=15$$, then $$5 \mathrm{a}+2 \mathrm{~b}$$ is :

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