The volume of a right triangular prism $ABC{A}_1{B}_1{C}_1$ is equal to 3. If the position vectors of the vertices of the base ABC are A(1, 0, 1), B(2, 0, 0) and C (0, 1, 0). The position vector of the vertex $A_1$ can be

Volume of prism = base area  height
 
$Ar\left(\Delta ABC\right)=\frac{1}{2}\left|\begin{array}{ccc}i& j& k\\ 1& 0& -1\\ 2& -1& 0\end{array}\right|$ 
$=\frac{1}{2}\sqrt{1^2+2^2+1^2}=\frac{\sqrt{6}}{2}$ 
volume = (base area) (height)
$\Rightarrow 3=\frac{\sqrt{6}}{2}\times h\Rightarrow h=\sqrt{6}$ units
Direction of edge $AA\text{'}=\overrightarrow{n}=\left|\begin{array}{ccc}i& j& k\\ 1& 0& -1\\ 2& -1& 0\end{array}\right|=⟨-1,-2,-1⟩$    
$\Rightarrow$ coordinates of $A\text{'}=(1,0,1)\pm \bcancel{\sqrt{6}}\left(\frac{-1,-2,-1}{\bcancel{\sqrt{6}}}\right)$  
= ( 0, – 2, 0) or (2, 2, 2)