If $(\alpha, \beta)$ is the orthocentre of the triangle $ABC$ with vertices $A(3, -7)$,$B(-1, 2)$,and $C(4, 5)$,then $9\alpha - 6\beta + 60$ is equal to:

What is the circumcenter of the triangle with vertices $(0, 0)$,$(3, 0)$,and $(0, 4)$?

If the vertices of a triangle are $(2, 1), (5, 2)$ and $(3, 4)$,then its circumcentre is

Let $A(1,0), B(6,2)$ and $C(\frac{3}{2}, 6)$ be the vertices of a triangle $ABC$. If $P$ is a point inside the triangle $ABC$ such that the triangles $APC, APB$ and $BPC$ have equal areas,then the length of the line segment $PQ,$ where $Q$ is the point $(-\frac{7}{6}, -\frac{1}{3})$ is

The medians of a $\Delta ABC$ are $9 \, cm, 12 \, cm$ and $15 \, cm$ respectively. Then the area of the triangle is.....$sq \, cm$.

The quadratic equation whose roots are the coordinates of the circumcentre of the triangle formed by the points $(-2,-1), (6,-1),$ and $(2,5)$ is