The absolute value of parameter t for which the area of the triangle whose vertices are A(-1,1,2);B(1,2,3) and C(1, 1, 1) is minimum.

$\begin{array}{l}\overrightarrow{\mathrm{AB}}=2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}},\overrightarrow{\mathrm{AC}}=(\mathrm{t}+1)\widehat{\mathrm{i}}+0\widehat{\mathrm{j}}-\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 2& 1& 1\\ \mathrm{t}+1& 0& -1\end{array}\right|\\ =-\widehat{\mathrm{i}}+(\mathrm{t}+3)\widehat{\mathrm{j}}-(\mathrm{t}+1)\widehat{\mathrm{k}}\\ =\sqrt{1+(\mathrm{t}+3{)}^2+(\mathrm{t}+1{)}^2}\\ =\sqrt{2{\mathrm{t}}^2+8\mathrm{t}+11}\end{array}$

$\begin{array}{r}\text{ Area of }\mathrm{\Delta ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}|\\ =\frac{1}{2}\sqrt{2{\mathrm{t}}^2+8\mathrm{t}+1}\end{array}$

$\begin{array}{l}\mathrm{let} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{\Delta }}^2=\frac{1}{4}\left(2{\mathrm{t}}^2+8\mathrm{t}+1\right)\\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{t}\right)=0\Rightarrow \mathrm{t}=-2\\ \mathrm{at} \mathrm{t}=-2,{\mathrm{f}}^{\prime \prime }\left(\mathrm{t}\right)>0\end{array}$

$\text{ So }\mathrm{\Delta }\text{ is minimum at }t=-2$