If the straight lines $x + 3y = 4,\,\,3x + y = 4$ and $x +y = 0$ form a triangle, then the triangle is

The equation of base $BC$ of an equilateral triangle is $3x + 4y = 1$ and vertex is $(-3,2),$ then the area of triangle is-

The triangle formed by the lines $x + y - 4 = 0,\,$ $3x + y = 4,$ $x + 3y = 4$ is

The area of triangle formed by the lines $x = 0,y = 0$ and $\frac{x}{a} + \frac{y}{b} = 1$, is

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

Let $\mathrm{A}(-2,-1), \mathrm{B}(1,0), \mathrm{C}(\alpha, \beta)$ and $\mathrm{D}(\gamma, \delta)$ be the vertices of a parallelogram $A B C D$. If the point $C$ lies on $2 x-y=5$ and the point $D$ lies on $3 x-2 y=6$, then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to_____.