A plane passes through the points A (1, 2, 3) B (2, 3, 1) and C(2, 4, 2). If O is the origin and P is   (2, -1, 1), then the projection of $\overrightarrow{OP}$on this plane is of length

The given points are A (1, 2, 3) B (2, 3, 1) and C(2, 4, 2).

            Equation of the plane passing through the above points is $\left|\begin{array}{ccc}x-1& y-2& z-3\\ 1& 1& -2\\ 1& 2& -1\end{array}\right|=0$

            Expand the above determinant, 

                                                $\begin{array}{c}(x-1)\left(3\right)-(y-2)\left(1\right)+(z-3)\left(1\right)=0\\ 3x-y+z-4=0\end{array}$

            Projection of O (0,0,0) on the above plane is (h,k,l)

                                 $\frac{h-0}{3}=\frac{k-0}{-1}=\frac{l-0}{1}=\frac{-(-4)}{9+1+1}=\frac{4}{11}$               

            Hence, the projection of the origin on the plane is $o^{\text{'}}$  =   $\left(\frac{12}{11},\frac{-4}{11},\frac{4}{11}\right)$

            Projection of P (2,-1,1) on the plane is (h,k,l)

                                       $\frac{h-2}{3}=\frac{k+1}{-1}=\frac{l-1}{l}=-\frac{(6+1+1-4)}{9+1+1}=\frac{-4}{11}$         

            Hence the projection of  on the plane is $P^{\text{'}}=\left(\frac{10}{11},\frac{-7}{11},\frac{7}{11}\right)$

            Therefore,$O\text{'}P\text{'}=\sqrt{\frac{4}{121}+\frac{9}{121}+\frac{9}{121}}=\sqrt{\frac{22}{121}}=\sqrt{\frac{2}{11}}$