Find 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(t, 1, 1) is minimum.

$\overline{AB}=2\widehat{i}+\widehat{j}+\widehat{k}, \overline{AC};=(t+1)\widehat{i}+0\widehat{j}-\widehat{k}$

$\overline{AB}\times \overline{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 1\\ t+1& 0& -1\end{array}\right|$

$=-\widehat{i}+(t+3)\widehat{j}-(t+1)\widehat{k}$

$=\sqrt{1+{(t+3)}^2+{(t+1)}^2}$

$=\sqrt{2t^2+8t+11}$

Area of $∆ABC=\frac{1}{2}\left|\overline{AB}\times \overline{AC}\right|$

$=\frac{1}{2}\sqrt{2t^2+8t+1}$

$Letf\left(t\right)={\Delta }^2=\frac{1}{4}(2t^2+8t+1)$

$f^{\text{'}}\left(t\right)=0\Rightarrow t=-2$

$Att=-2, f\text{'}\text{'}\left(t\right)>0$

So $∆$ is minimum  at  t = - 2