What is the angle between the diagonals of the parallelogram formed by the lines $ℓx + my + n = 0$,$ℓx + my + n' = 0$,$mx + ℓy + n = 0$,and $mx + ℓy + n' = 0$?

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 $ABCD$. If $AB = \sqrt{10}$ and the points $A$ and $C$ lie on the line $3y = 2x + 1$,then $2(\alpha + \beta + \gamma + \delta)$ is equal to

Let $A \equiv (3, 2)$ and $B \equiv (5, 1)$. An equilateral triangle $ABP$ is constructed on the side of $AB$ remote from the origin. The orthocentre of triangle $ABP$ is:

If the points $(k, 3), (2, k),$ and $(-k, 3)$ are collinear,then the values of $k$ are

The perimeter of a triangle is $16 \text{ cm}$, one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$, then the triangle is:

In a $\triangle ABC$,medians $AD$ and $BE$ are drawn. If $AD = 4$,$\angle DAB = \frac{\pi}{6}$,and $\angle ABE = \frac{\pi}{3}$,then the area of the $\triangle ABC$ is