$$\text { Line } L \Rightarrow \frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}=\mu$$

Point $$M(\mu, \mu+3,-\mu+1)$$

Direction vector of $$Q M=(\mu-3, \mu+6,-\mu)$$

Direction vector of line $$L=(1,1,-1)$$

Both direction vector are perpendicular so

$$\begin{aligned} & 1 \times(\mu-3)+1 \times(\mu+6)-1 \times(-\mu)=0 \\ & \mu=-1 \end{aligned}$$

Point $$M(-1,2,2)$$ midpoint of $$P Q$$

So point $$P(\alpha, \beta, \gamma)=(-5,7,3)$$

Area of $$\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{Q R}|$$

$$\begin{aligned} & \overrightarrow{P Q}=(8,-10,-2) \\ & \overrightarrow{Q R}=(-1,8,-2) \\ & \overrightarrow{P Q} \times \overrightarrow{Q R}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 8 & -10 & -2 \\ -1 & 8 & -2 \end{array}\right|=36 \hat{i}+18 \hat{j}+54 \hat{k} \end{aligned}$$

Area of $$\triangle P Q R=\lambda$$

$$\lambda=\frac{1}{2} \sqrt{(36)^2+(18)^2+(54)^2}$$

$$\begin{aligned} & \lambda^2=\frac{1}{4}(4536) \\ & \lambda^2=1134 \\ & \lambda^2=14 K \text { (given) } \\ & 14 K=1134 \\ & K=\frac{1134}{14}=81 \\ & K=81 \end{aligned}$$