If the area of the triangle formed by the points $\left(2\mathrm{a},\mathrm{b}\right)$ $\left(\mathrm{a}+\mathrm{b},2\mathrm{b}+\mathrm{a}\right),$ and $\left(2\mathrm{b},2\mathrm{a}\right)$ is 2 sq. units, then the area of the triangle whose vertices and $\left(\mathrm{a}+\mathrm{b},\mathrm{a}-\mathrm{b}\right),\left(3\mathrm{b}-\mathrm{a},\mathrm{b}+3\mathrm{a}\right)$ and $\left(3\mathrm{a}-\mathrm{b},3\mathrm{b}-\mathrm{a}\right)$ will be