If the area of the triangle with vertices \((1,2,0)\), \((1,0,2)\) and \((0, x, 1)\) is \(\sqrt{6}\) square units, then the value of \(x\) is

\(\begin{aligned} & \text { Let } \mathrm{A} \equiv(1,2,0), \mathrm{B} \equiv(1,0,2) \text { and } \mathrm{C} \equiv(0, x, 1) \\ & \therefore \quad \overline{\mathrm{AB}}=-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overline{\mathrm{AC}}=-\hat{\mathrm{i}}+(x-2) \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \therefore \quad \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|=\sqrt{6} \\ & |\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & -2 & 2 \\ -1 & x-2 & 1\end{array}\right| \\ & =\hat{\mathrm{i}}[-2-2(x-2)]-\hat{\mathrm{j}}(0+2)+\hat{\mathrm{k}}(0-2) \\ & =(2-2 x) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \therefore \quad \frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|=\sqrt{6} \\ & \Rightarrow \frac{1}{2} \sqrt{(2-2 x)^2+4+4}=\sqrt[6]{6} \\ & \Rightarrow(2-2 x)^2=16 \\ & \Rightarrow 4-8 x+4 x^2=16 \\ & \Rightarrow x^2-2 x+3=0 \\ & \Rightarrow(x-3)(x+1)=0 \\ & \Rightarrow x=3 \text { or }-1 \\ & \end{aligned}\)