Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$ and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$ and $\vec{r} \cdot(\vec{b}-\vec{c})=0$, then $\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$ is equal to __________.

<p>$$\begin{aligned} & \vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k} \\ & \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k} \\ & \vec{c}=17 \hat{i}-2 \hat{j}+\hat{k} \\ & \vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a} \\ & (\vec{r}-(\vec{b}+\vec{c})) \times \vec{a}=0 \\ & \Rightarrow \vec{r}=(\vec{b}+\vec{c})+\lambda \vec{a} \\ & \vec{r}=(20 \hat{i}+5 \hat{j}-12 \hat{k})+\lambda(9 \hat{i}-13 \hat{j}+25 \hat{k}) \\ & =(20+9 \lambda) \hat{i}+(5-13 \lambda) \hat{j}+(25 \lambda-12) \hat{k} \end{aligned}$$</p> <p>Now $\vec{r} \cdot(\vec{b}-\vec{c})=0$</p> <p>$\vec{r} \cdot(-14 \hat{i}+9 \hat{j}-14 \hat{k})=0$</p> <p>Now</p> <p>$$\begin{aligned} & -14(20+9 \lambda)+9(5-13 \lambda)-14(25 \lambda-12)=0 \\ & -593 \lambda-67=0 \\ & \lambda=-\frac{67}{593} \\ & \therefore \vec{r}=(\vec{b}+\vec{c})-\frac{67}{593} \vec{a} \\ & \frac{|593 \vec{r}+67 \vec{a}|^2}{|593|^2}=|\vec{b}+\vec{c}|^2=|20 \hat{i}+5 \hat{j}-12 \hat{k}|^2 \\ & =569 \end{aligned}$$</p>