If vertices of a triangle are A(1,- 1, Z), B(2,0,- 1) and C(0,2,1), then the area of a triangle is

# If vertices of a triangle are A(1,- 1, Z), B(2,0,- 1) and C(0,2,1), then the area of a triangle is

1. A

$\sqrt{6}$

2. B

$2\sqrt{6}$

3. C

$3\sqrt{6}$

4. D

$4\sqrt{6}$

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### Solution:

Now,

$\begin{array}{l}{\mathrm{\Delta }}_{\mathrm{xy}}=\frac{1}{2}\left|\left|\begin{array}{ccc}1& -1& 1\\ 2& 0& 1\\ 0& 2& 1\end{array}\right|\right|\\ =\frac{1}{2}|1\left(0-2\right)+1\left(2-0\right)+1\left(4-0\right)|\\ =\frac{1}{2}|-2+2+4|=2\end{array}$

$\begin{array}{l}{\mathrm{\Delta }}_{\mathrm{yz}}=\frac{1}{2}\left|\left|\begin{array}{ccc}-1& 2& 1\\ 0& -1& 1\\ 2& 1& 1\end{array}\right|\right|\\ =\frac{1}{2}|\left[-1\left(-1-1\right)-2\left(0-2\right)+1\left(0+2\right)\right]|\\ =\frac{1}{2}|\left[2+4+2\right]|=4\end{array}$

$\begin{array}{l}{\mathrm{\Delta }}_{\mathrm{zx}}=\frac{1}{2}\left|\left|\begin{array}{ccc}2& 1& 1\\ -1& 2& 1\\ 1& 0& 1\end{array}\right|\right|\\ =\frac{1}{2}|\left[2\left(2-0\right)-1\left(-1-1\right)+1\left(0-2\right)\right]|\\ =\frac{1}{2}|\left[4+2-2\right]|=2\end{array}$

$\therefore$ Area of triangle,

$\begin{array}{l}\mathrm{\Delta }=\sqrt{{\mathrm{\Delta }}_{\mathrm{xy}}^{2}+{\mathrm{\Delta }}_{\mathrm{yz}}^{2}+{\mathrm{\Delta }}_{\mathrm{zx}}^{2}}\\ =\sqrt{{2}^{2}+{4}^{2}+{2}^{2}}=\sqrt{24}=2\sqrt{6}\end{array}$

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