1. Given vectors

$\vec a = \hat i + 2\hat j - 3\hat k = (1,2,-3), \qquad \vec b = 2\hat i - 3\hat j + 5\hat k = (2,-3,5)$

Let

$\vec r = x\hat i + y\hat j + z\hat k = (x,y,z).$

2. Use the condition

$\vec r \times \vec a = \vec b \times \vec r$

Since

$\vec b \times \vec r = -(\vec r \times \vec b),$

we get

$\vec r \times \vec a = -\vec r \times \vec b$

so

$\vec r \times (\vec a + \vec b)=0.$

Now,

$\vec a + \vec b = (1+2,\,2-3,\,-3+5)=(3,-1,2).$

Hence

$\vec r \times (3,-1,2)=0.$

Therefore, $\vec r$ is parallel to $(3,-1,2)$, so

$\vec r = \lambda(3,-1,2)=(3\lambda,-\lambda,2\lambda).$

3. Use the first scalar condition

Given

$\vec r \cdot (\alpha \hat i + 2\hat j + \hat k)=3$

that is,

$\vec r \cdot (\alpha,2,1)=3.$

Substitute $\vec r=(3\lambda,-\lambda,2\lambda)$:

$3\lambda\alpha + (-\lambda)(2) + (2\lambda)(1)=3.$

So,

$3\lambda\alpha -2\lambda +2\lambda = 3$

$3\lambda\alpha=3$

$\lambda\alpha =1. \qquad (1)$

4. Use the second scalar condition

Given

$\vec r \cdot (2\hat i + 5\hat j - \alpha \hat k)=-1$

that is,

$\vec r \cdot (2,5,-\alpha)=-1.$

Substitute $\vec r=(3\lambda,-\lambda,2\lambda)$:

$3\lambda(2)+(-\lambda)(5)+2\lambda(-\alpha)=-1$

$6\lambda-5\lambda-2\alpha\lambda=-1$

$\lambda - 2\alpha\lambda = -1.$

Using (1), $\alpha\lambda=1$:

$\lambda -2 = -1$

$\lambda=1.$

Then from $\alpha\lambda=1$,

$\alpha=1.$

5. Find $|\vec r|^2$

Since $\lambda=1$,

$\vec r=(3,-1,2).$

Therefore,

$|\vec r|^2 = 3^2+(-1)^2+2^2 = 9+1+4=14.$

6. Compute the required value

$\alpha + |\vec r|^2 = 1+14=15.$

So the correct option is

$\boxed{15}$

which is Option D.