If $$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$$, $$\vec{b} = 3(\hat{i} - \hat{j} + \hat{k})$$ and $$\vec{c}$$ be the vector such that $$\vec{a} \times \vec{c} = \vec{b}$$ and $$\vec{a} \cdot \vec{c} = 3$$, then $$\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c})$$ is equal to

We are given
$$\vec{a}=1\,\hat{i}+2\,\hat{j}+1\,\hat{k},\qquad \vec{b}=3(\hat{i}-\hat{j}+\hat{k})=3\,\hat{i}-3\,\hat{j}+3\,\hat{k}.$$

Let $$\vec{c}=x\,\hat{i}+y\,\hat{j}+z\,\hat{k}.$$
The two conditions on $$\vec{c}$$ are

1. $$\vec{a}\times\vec{c}=\vec{b}$$

2. $$\vec{a}\cdot\vec{c}=3$$

Case 1: Solve $$\vec{a}\times\vec{c}=\vec{b}.$$ Using the determinant rule for a cross product,

$$\vec{a}\times\vec{c}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 1&2&1\\ x&y&z \end{vmatrix} =\hat{i}(2z-y)-\hat{j}(z-x)+\hat{k}(y-2x).$$

Equating components with $$\vec{b}=(3,\,-3,\,3),$$ we obtain

$$2z-y=3\quad -(1)$$

$$-(z-x)=-3\;\Longrightarrow\; z-x=3\quad -(2)$$

$$y-2x=3\quad -(3)$$

Case 2: Solve $$\vec{a}\cdot\vec{c}=3.$$ The dot product gives

$$1\cdot x+2\cdot y+1\cdot z=3\quad -(4)$$

From $$(2)$$ we get $$x=z-3.$$
From $$(1)$$ we get $$y=2z-3.$$

Substitute these in $$(4):$$

$$(z-3)+2(2z-3)+z=3$$

$$z-3+4z-6+z=3$$

$$6z-9=3$$

$$6z=12\;\Longrightarrow\; z=2.$$

Hence

$$x=z-3=2-3=-1,\qquad y=2z-3=4-3=1.$$

Therefore $$\vec{c}=-1\,\hat{i}+1\,\hat{j}+2\,\hat{k}.$$

Case 3: Evaluate the required expression $$\vec{a}\cdot\bigl((\vec{c}\times\vec{b})-\vec{b}-\vec{c}\bigr).$$

First compute $$\vec{c}\times\vec{b}:$$

$$\vec{c}\times\vec{b}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ -1&1&2\\ 3&-3&3 \end{vmatrix} =\hat{i}(1\cdot3-2\cdot(-3)) -\hat{j}((-1)\cdot3-2\cdot3) +\hat{k}((-1)(-3)-1\cdot3)$$

$$=\hat{i}(3+6)-\hat{j}(-3-6)+\hat{k}(3-3) =9\,\hat{i}+9\,\hat{j}+0\,\hat{k}.$$

Now form the vector inside the parentheses:

$$(\vec{c}\times\vec{b})-\vec{b}-\vec{c} =(9,9,0)-(3,-3,3)-(-1,1,2)$$

$$=(9-3+1,\;9+3-1,\;0-3-2) =(7,\;11,\;-5).$$

Finally take the dot product with $$\vec{a}=(1,2,1)\!:\!$$

$$\vec{a}\cdot(7,11,-5)=1\cdot7+2\cdot11+1\cdot(-5)=7+22-5=24.$$

Hence the value of $$\vec{a}\cdot\bigl((\vec{c}\times\vec{b})-\vec{b}-\vec{c}\bigr)=24.$$

Therefore, the correct option is Option B (24).