the three vertices of a parallelogram …

Diagonals of a parallelogram bisect each other.
So E is mid point of AC, So Coordinates of E are {tex}E(\frac{3-6}{2},\frac{-4+2}{2}){/tex} or {tex} E ( \frac{-3}{2},-1)....(1){/tex}
Again E is the mid point of BD so coordinates of E are:
{tex}E ( \frac{x-1}{2}, \frac{y-3}{2})....(2){/tex}

from (1) and (2 ) we get 
{tex}\frac{x-1}{2}{/tex} = -{tex}\frac{3}{2}{/tex}
x - 1 = -3
x = -2
{tex}\frac{y-3}{2}{/tex} = -1
y - 3 = -2
y = 1
Hence the coordinates of fourth vertex D are (-2, 1)
Now area of {tex}\triangle{/tex}ABC,
{tex}= \frac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( y _ { 1 } - y _ { 2 } \right) \right]{/tex}
{tex}= \frac { 1 } { 2 } [ 3 ( - 3 - 2 ) - 1 ( 2 + 4 ) - 6 ( - 4 + 3 ) ]{/tex}
{tex}= \frac { 1 } { 2 } [ - 15 - 6 + 6 ]{/tex}
{tex}= \frac { 1 } { 2 } \times ( - 15 ){/tex}
{tex}= - \frac { 15 } { 2 }{/tex}
{tex}= \frac { 15 } { 2 }{/tex} sq. units
Since, diagonal divides parallelogram into two equal parts
So, Area of parallelogram ABCD
= 2 {tex}\times{/tex} Area of {tex}\triangle{/tex} ABC
{tex}= 2 \times \frac { 15 } { 2 }{/tex}= 15 sq.units.