A plane passes through the points

Solution:

$\mathrm{A}(1,2,3), \mathrm{B}(2,3,1), \mathrm{C}(2,4,2), \mathrm{O}(0,0,0)$

Equation of plane passing through $A, B, C$ will be

$\left|\begin{array}{lll}x-1 & y-2 & z-3 \\ 2-1 & 3-2 & 1-3 \\ 2-1 & 4-2 & 2-3\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}x-1 & y-2 & z-3 \\ 1 & 1 & -2 \\ 1 & 2 & -1\end{array}\right|=0$

$\Rightarrow(x-1)(-1+4)-(y-2)(-1+2)+(z-3)(2-1)=0$

$\Rightarrow(x-1)(3)-(y-2)(1)+(z-3)(1)=0$

$\Rightarrow 3 x-3-y+2+z-3=0$

$\Rightarrow 3 x-u+z-4=0$. is the reauired plane.

Now, given $O(0,0,0) \& P(2,-1,1)$

Plane is $3 x-y+z-4=0$

$\mathrm{O}^{\prime} \& \mathrm{P}^{\prime}$ are foot of perpendiculars.

for $\mathrm{O}^{\prime}$

$\frac{x-0}{3}=\frac{y-0}{-1}=\frac{z-0}{1}=\frac{-(0-0+0-4)}{9+1+1}$

$\frac{x}{3}=\frac{y}{-1}=\frac{z}{1}=\frac{4}{11}$

$\Rightarrow O^{\prime}\left(\frac{12}{11}, \frac{-4}{11}, \frac{4}{11}\right)$

for $\mathrm{P}^{\prime}$

$\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-1}{1}=\frac{-(3(2)-(-1)+1-4)}{9+1+1}$

$\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-1}{1}=\left(\frac{-4}{11}\right)$

$P^{\prime}\left(\frac{-12}{11}+2, \frac{4}{11}-1, \frac{-4}{11}+1\right)$

$\Rightarrow P^{\prime}\left(\frac{10}{11}, \frac{-7}{11}, \frac{7}{11}\right)$

$O^{\prime} P^{\prime}=\sqrt{\left(\frac{10}{11}-\frac{12}{11}\right)^{2}+\left(\frac{-7}{11}+\frac{4}{11}\right)^{2}+\left(\frac{7}{11}-\frac{4}{11}\right)^{2}}$

$\Rightarrow O^{\prime} P^{\prime}=\frac{1}{11} \sqrt{4+9+9}$

$\Rightarrow \quad O^{\prime} P^{\prime}=\frac{\sqrt{22}}{11}$

$\Rightarrow \quad O^{\prime} P^{\prime}=\frac{\sqrt{2} \times \sqrt{11}}{11}$

$\Rightarrow O^{\prime} P^{\prime}=\sqrt{\frac{2}{11}}$