Triangle defined by one point III

Let $A(a,0,0), B(0,b,0), C(0,0,c)$ denote vertices of the triangle. The centroid divides each median into parts in the ratio 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex. That same ratio holds for the projection of the median onto an axis on which the median vertex lies. Therefore:

$a = 3 \times 1 = 3, b = 2 \times 3 = 6, c = 3 \times 3 = 9$ .

By de Gua's theorem

$\triangle ABC = \sqrt { { \left( \dfrac { 1 }{ 2 } 3\cdot 6 \right) }^{ 2 }+{ \left( \dfrac { 1 }{ 2 } 6\cdot 9 \right) }^{ 2 }+{ \left( \dfrac { 1 }{ 2 } 9\cdot 3 \right) }^{ 2 } } =\dfrac { 63 }{ 2 } =\boxed{31.5}$

I'll use some knowledge about vector.

First, centroid can be found by calculating the average of vertice's coordinate. So, the coordinate of the three vertice are $(3,0,0),(0,6,0),(0,0,9)$

Then, let's go through vector!, the two vectors that compose two sides of the triangle are $\vec{u}=(-3,6,0)$ and $\vec{v}(-3,0,9)$

Hence, the area of the triangle is $\dfrac{1}{2} \left| \vec{u} \times \vec{v} \right|=\dfrac{9}{2} \left| 6\hat{i}+3\hat{j}+2\hat{k} \right|$

$=\dfrac{9}{2} \times \sqrt{49}=\dfrac{63}{2}=\boxed{31.5}$