lf there are three points A(2,3, 4), B(- 1,2, - 3) and C(-4,1, - 10) in a space, then they are

Given, points are A(2,3,4), B(-1,2, -3) and C(-4,1, -10).
Now, determine area of triangle
Area of triangle  $\mathrm{\Delta }=\sqrt{{\mathrm{\Delta }}_{\mathrm{xy}}^2+{\mathrm{\Delta }}_{\mathrm{\gamma z}}^2+{\mathrm{\Delta }}_{\mathrm{zx}}^2}$

Where, ${\mathrm{\Delta }}_{\mathrm{xy}}=\frac{1}{2}\left|\left|\begin{array}{l}{\mathrm{x}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{x}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{x}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|\right|$ , ${\mathrm{\Delta }}_{\mathrm{yz}}=\frac{1}{2}\left|\left|\begin{array}{l}{\mathrm{y}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{z}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{y}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{z}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{y}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{z}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|\right|$  and ${\mathrm{\Delta }}_{\mathrm{zx}}=\frac{1}{2}\left|\left|\begin{array}{l}{\mathrm{z}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{z}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{z}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|\right|$

now,

$\begin{array}{c}{\mathrm{\Delta }}_{\mathrm{xy}}=\frac{1}{2}\left|\left|\begin{array}{ccc}2& 3& 1\\ -1& 2& 1\\ -4& 1& 1\end{array}\right|\right|\\ =\frac{1}{2}\left|\left[2(2-1)-3(-1+4)]+1(-1+8)\right]\right|\\ =\frac{1}{2}\left|\right[2-9+7\left]\right|=0\end{array}$

$\begin{array}{c}{\mathrm{\Delta }}_{\mathrm{yz}}=\frac{1}{2}\left|\left|\begin{array}{ccc}3& 4& 1\\ 2& -3& 1\\ 1& -10& 1\end{array}\right|\right|\\ =\frac{1}{2}\left|\left[3(-3+10)-4(2-1)+1(-20+3)\right]\right|\\ =\frac{1}{2}|21-4-17|=0\end{array}$

$\begin{array}{c}{\mathrm{\Delta }}_{\mathrm{zx}}=\frac{1}{2}\left|\left|\begin{array}{ccc}4& 2& 1\\ -3& -1& 1\\ -10& -4& 1\end{array}\right|\right|\\ =\frac{1}{2}\left|\left[4(-1+4)-2(-3+10)+1(12-10)\right]\right|\\ =\frac{1}{2}\left|\right[12-14+2\left]\right|=0\\ \mathrm{\Delta }=\sqrt{0^2+0^2+0^2}=0\end{array}$

Hence, given points are collinear