Let the position vectors of two points P and Q

Solution:

$\mathrm{P}(3,-1,2)$

$\mathrm{Q}(1,2,-4)$

$\overline{\operatorname{PR}} \| 4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$\overline{\mathrm{QS}} \|-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

dr's of normal to the plane containing $\mathrm{P}, \mathrm{T} \& \mathrm{Q}$ will be proportional to :

$\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & -1 & 2 \\ -2 & 1 & -2\end{array}\right|$

$\therefore \quad \frac{\ell}{0}=\frac{\mathrm{m}}{4}=\frac{\mathrm{n}}{2}$

For point, $\mathrm{T}: \overrightarrow{\mathrm{PT}}=\frac{\mathrm{x}-3}{4}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\lambda$

$\overrightarrow{\mathrm{QT}}=\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}+4}{-2}=\mu$

$\mathrm{T}:(4 \lambda+3,-\lambda-1,2 \lambda+2)$

$\cong(2 \mu+1, \mu+2,-2 \mu-4)$

$4 \lambda+3=-2 \mu+1 \Rightarrow 2 \lambda+\mu=-1$

$\lambda+\mu=-3 \quad \Rightarrow \quad \lambda=2$

$\& \mu=-5 \quad \lambda+\mu=-3 \quad \Rightarrow \quad \lambda=2$

So point T : $(11,-3,6)$

$\overrightarrow{\mathrm{OA}}=(11 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \pm\left(\frac{2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{5}}\right) \sqrt{5}$

$\overrightarrow{\mathrm{OA}}=(11 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \pm(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$

$\overrightarrow{\mathrm{OA}}=11 \hat{\mathrm{i}}-\hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

Or

$9 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$

$|\overrightarrow{\mathrm{OA}}|=\sqrt{121+1+49}=\sqrt{171}$

Or

$\sqrt{81+25+25}=\sqrt{131}$