$A(1, -2), B(-2, 3), C(-1, -3)$ are the vertices of a triangle $ABC$. $L_1$ is the perpendicular drawn from $A$ to $BC$ and $L_2$ is the perpendicular bisector of $AB$. If $(l, m)$ is the point of intersection of $L_1$ and $L_2$, then $26m - 3 =$ (in $l$)

In a triangle $PQR$,the coordinates of the points $P$ and $Q$ are $(-2, 4)$ and $(4, -2)$ respectively. If the equation of the perpendicular bisector of $PR$ is $2x - y + 2 = 0$,then the centre of the circumcircle of the $\Delta PQR$ is

Let $A(6,8)$,$B(10 \cos \alpha, -10 \sin \alpha)$ and $C(-10 \sin \alpha, 10 \cos \alpha)$ be the vertices of a triangle. If $L(a, 9)$ and $G(h, k)$ are its orthocenter and centroid respectively,then $(5a - 3h + 6k + 100 \sin 2\alpha)$ is equal to . . . . . .

If $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$ are the vertices of a triangle with side lengths $a, b, c$ opposite to vertices $A, B, C$ respectively,then the excentre with respect to vertex $B$ is:

Two vertices of a triangle $ABC$ are $A(3, -1)$ and $B(-2, 3)$,and its orthocentre is $P(1, 1)$. If the coordinates of the point $C$ are $(\alpha, \beta)$ and the centre of the circle circumscribing the triangle $PAB$ is $(h, k)$,then the value of $(\alpha + \beta) + 2(h + k)$ equals :

The orthocentre of the triangle with vertices $(-2, -6)$,$(-2, 4)$ and $(1, 3)$ is