Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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