Let $\vec a = - \hat i - \hat k$, $\vec b = - \hat i + \hat j$ and $\vec c = \hat i + 2\hat j + 3\hat k$ be the three given vectors. If $\vec r$is a vector such that $\vec r \times \vec b = \vec c \times \vec b$ and $\vec r.\vec a = 0$, then find the value of $\vec r.\vec b$ is
A) 3
B) 6
C) 9
D) 12

Hint: Let the vector $r$ is $(x,y,z)$ and substitute it in two given equations. After substituting the value of the vector $r$, we get simple algebraic equations in the form of $x$, $y$ and $z$. After solving these equations, we get values of $x$, $y$ and $z$ and then we have a vector $r$. Finally, we find vector $r$ and vector $b$ is already given then calculate the value of $\vec r.\vec b$.

Complete step-by-step answer:
Let $\vec r = x\hat i + y\hat j + z\hat k$
Given $\vec r \times \vec b = \vec c \times \vec b$ and $\vec r.\vec a = 0$.
$\vec r \times \vec b = \vec c \times \vec b$
$(x\hat i + y\hat j + z\hat k) \times ( - \hat i + \hat j) = (\hat i + 2\hat j + 3\hat k) \times ( - \hat i + \hat j)$
After cross multiplication on both side we get
$ - z\hat i - z\hat j + (x + y)\hat k = - 3\hat i - 3\hat j + 3\hat k$
From above eqn. we get
 $z = 3$ and $x + y = 3$-(i)
Now expand eqn. $\vec r.\vec a = 0$ we get
$ - x - z = 0$ means $x = - z = - 3$
After putting the value of $x$ in eqn. (i),
 $y = 3 - x$ then $y = 3 - ( - 3) = 6$
Now $\vec r = - 3\hat i + 6\hat j + 3\hat k$
We have to find $\vec r.\vec b$
$\vec r.\vec b = ( - 1)( - 3) + 6.1 + 3.0 = 3 + 6 = 9$

Hence, the correct answer is option C.

Note: Cross product of two vectors gives a vector and dot product of two vectors gives us a simple number. Two vectors are equal if and only if each component of one vector is equal to the same component of second vector e.g. value of $\hat i$ component of first vector is equal to the value of $\hat i$ component of second vector and same of all other components. The vector result of cross product of two vectors is perpendicular to the both vectors.