Let the position vectors of two points P and Q be 3 i- j+2 k and i+2 j-4 k, respectively. Let R and S be two points such that the direction ratios of…

$$P(3,-1,2)$$P(3,−1,2)

$$Q(1,2,-4)$$Q(1,2,−4)

$$\vec{PR}\|4\hat{i}-\hat{j}+2\hat{k}$$PR∥4i^−j^​+2k^

$$\vec{QS}\|-2\hat{i}+\hat{j}-2\hat{k}$$QS​∥−2i^+j^​−2k^

Dr's of normal to the plane containing $P,T$P,T and $Q$Q will be proportional to :

$$|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 2 \\ -2 & 1 & -2\end{matrix}|=4\hat{j}+2\hat{k}$$∣i^4−2​j^​−11​k^2−2​∣=4j^​+2k^

$$\therefore \frac{\ell }{0}=\frac{m}{4}=\frac{n}{2}$$∴0ℓ​=4m​=2n​

For point, $T$T : $\vec{PT}=\frac{x-3}{4}=\frac{y+1}{-1}=\frac{z-2}{2}=\lambda$PT=4x−3​=−1y+1​=2z−2​=λ

$$\vec{QT}=\frac{x-1}{-2}=\frac{y-2}{1}=\frac{z+4}{-2}=\mu$$QT​=−2x−1​=1y−2​=−2z+4​=μ

$$T\equiv (4\lambda +3,-\lambda -1,2\lambda +2)$$T≡(4λ+3,−λ−1,2λ+2)

$$Q\equiv (2\mu +1,\mu +2,-2\mu -4)$$Q≡(2μ+1,μ+2,−2μ−4)

$$4\lambda +3=-2\mu +1\Rightarrow 2\lambda +\mu =-1$$4λ+3=−2μ+1⇒2λ+μ=−1

$$\lambda +\mu =-3\Rightarrow \lambda =2$$λ+μ=−3⇒λ=2

and $\mu =-5,\lambda +\mu =-3\Rightarrow \lambda =2$μ=−5,λ+μ=−3⇒λ=2

So point $T:(11,-3,6)$T:(11,−3,6)

$$\vec{OA}=(11\hat{i}-3\hat{j}+6\hat{k})\pm (\frac{2\hat{j}+\hat{k}}{\sqrt{5}})\sqrt{5}$$OA=(11i^−3j^​+6k^)±(5​2j^​+k^​)5​

$$\vec{OA}=(11\hat{i}-3\hat{j}+6\hat{k})\pm (2\hat{j}+\hat{k})$$OA=(11i^−3j^​+6k^)±(2j^​+k^)

$\vec{OA}=11\hat{i}-\hat{j}+7\hat{k}$OA=11i^−j^​+7k^ or $9\hat{i}-5\hat{j}+5\hat{k}$9i^−5j^​+5k^

$|\vec{OA}|=\sqrt{121+1+49}=\sqrt{171}$∣OA∣=121+1+49​=171​ or $\sqrt{81+25+25}=\sqrt{131}$81+25+25​=131​