Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2 \sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{O A} \cdot \overrightarrow{O B}$ is equal to

<p>$L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$</p> <p>Point $A(3 \lambda+6,2 \lambda+7,7-2 \lambda)$</p> <p>$B(3 \mu+6,2 \mu+7,7-2 \mu)$</p> <p>Let $P(3 k+6,2 k+7,7-2 k)$ be foot of perpendicular from $P^{\prime}(1,2,3)$</p> <p>$$\begin{aligned} & \therefore \quad P P^{\prime}\langle 3,2,-2\rangle=0 \\ & 3(3 k+5)+(2 k+5) 2+(4-2 k)(-2)=0 \\ & 9 k+15+4 k+10-8+4 k=0 \\ & 17 k+17=0 \\ & \Rightarrow k=-1 \qquad \therefore \quad P(3,5,9)\\ & |\overrightarrow{A P}|=2 \sqrt{17} \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow(3 \lambda+3)^2+(2 \lambda+2)^2+(-2-2 \lambda)^2=17 \times 4 \\ & =17(\lambda+1)^2=17 \times 4 \\ & \Rightarrow \lambda+1= \pm 2 \Rightarrow \lambda=1 \text { or } \lambda=-3 \\ & \therefore A(9,9,5), B(-3,1,13) \\ & \overrightarrow{O A} \cdot \overrightarrow{O B}=(9 \hat{i}+9 \hat{j}+5 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+13 \hat{k}) \\ & =-27+9+65=47 \end{aligned}$$</p>