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 ________

We are given $$\vec{a}=(9,-13,25)$$, $$\vec{b}=(3,7,-13)$$, and $$\vec{c}=(17,-2,1)$$, and we wish to find $$\frac{|593\vec{r}+67\vec{a}|^2}{593^2}$$ under the conditions $$\vec{r}\times\vec{a}=(\vec{b}+\vec{c})\times\vec{a}$$ and $$\vec{r}\cdot(\vec{b}-\vec{c})=0$$.

Since $$\vec{r}\times\vec{a}=(\vec{b}+\vec{c})\times\vec{a}$$ implies $$(\vec{r}-\vec{b}-\vec{c})\times\vec{a}=0$$, the vector $$\vec{r}$$ must have the form $$\vec{r}=\vec{b}+\vec{c}+\lambda\vec{a}$$ for some scalar $$\lambda$$. Noting that $$\vec{b}+\vec{c}=(20,5,-12)$$, we write $$\vec{r}=(20+9\lambda,5-13\lambda,-12+25\lambda)$$.

Next, using the perpendicularity condition $$\vec{r}\cdot(\vec{b}-\vec{c})=0$$ with $$\vec{b}-\vec{c}=(-14,9,-14)$$, we substitute the coordinates of $$\vec{r}$$ to get

$$ (20+9\lambda)(-14)+(5-13\lambda)(9)+(-12+25\lambda)(-14)=0\,. $$

Expanding and combining like terms yields

$$ -280-126\lambda+45-117\lambda+168-350\lambda=0 $$

which simplifies to $$-67-593\lambda=0$$ and hence $$\lambda=-\frac{67}{593}\,. $$

We then compute $$593\vec{r}+67\vec{a}$$ by first observing

$$ 593\vec{r}=593(\vec{b}+\vec{c}+\lambda\vec{a})=593(\vec{b}+\vec{c})+593\lambda\vec{a}=593(\vec{b}+\vec{c})-67\vec{a}\,, $$

so that $$593\vec{r}+67\vec{a}=593(\vec{b}+\vec{c})\,. $$

Since $$\vec{b}+\vec{c}=(20,5,-12)$$, its squared length is $$|\vec{b}+\vec{c}|^2=20^2+5^2+(-12)^2=400+25+144=569\,. $$ Therefore,

$$ |593(\vec{b}+\vec{c})|^2=593^2\times569\,, $$

and hence

$$ \frac{|593\vec{r}+67\vec{a}|^2}{593^2}=569. $$

Thus, the final answer is 569.