A triangle has vertices at `(1,1,1),(2,2,2),(1,1,y)` and area of the triangle is `cosec (pi)/(4)` square units then value of `y` is /are

A triangle has vertices (1,1,1),(2,2,2),(1,1,y) and its area equals to cosec(pi/4) sq.units. The sum of the possible values of y is

area of triangle whose vertices are (1,-1),(2,1) and (4,5) is :

Area of the triangle having vertices (1,2,3),(2,-1,1) and (1,2,4) in square units is

If the area of the triangle with vertices x,0,1,1 and 0,2 1, 4 sq units,then the value of x is :

the area of triangle whose vertices are (1,2,3),(2,5-1) and (-1,1,2) is

Vertices of a triangle are A(3, 1, 2), B(1, -1, 2) and C(2, 1, 1) . Area of the triangle will be :

Find the area of the triangle with vertices (3,8),(-4,2) and (5,1) .

Area of triangle whose vertices are (a,a^2),(b,b^2),(c,c^2) is 1/2 . and area of another triangle whose vertices are (p,p^2),(q,q^2) and (r,r^2) is 4, then the value of |((1+ap)^2,(1+bp)^2,(1+cp)^2),((1+aq)^2,(1+bp)^2,(1+cq)^2),((1+ar)^2,(1+br)^2,(1+cr)^2)| is (A) 2 (B) 4 (C) 8 (D) 16

Find the area of the triangle if vertices are (1,1), (2,2) and (0,1)