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Step-by-Step Solution
Step 1: Identify the given vectors and conditions
Given:
$ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \implies \vec{a} = (1,\,2,\,1). $
$ \vec{b} = 3(\hat{i} - \hat{j} + \hat{k}) \implies \vec{b} = (3,\, -3,\, 3). $
Let $ \vec{c} = (x,\, y,\, z). $
We know:
1. $ \vec{a} \times \vec{c} = \vec{b}. $
2. $ \vec{a} \cdot \vec{c} = 3. $
We want to find:
$ \vec{a} \cdot\bigl[(\vec{c} \times \vec{b}) - \vec{b} - \vec{c}\bigr]. $
Step 2: Write the cross product equation $\vec{a} \times \vec{c} = \vec{b}$
The cross product $ \vec{a} \times \vec{c} $ is:
$$
\vec{a} \times \vec{c}
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
x & y & z
\end{vmatrix}
= \bigl(2z - y,\; x - z,\; y - 2x\bigr).
$$
This must be equal to $ \vec{b} = (3,\, -3,\, 3). $ Hence we get the system:
$$
2z - y = 3,\quad x - z = -3,\quad y - 2x = 3.
$$
Step 3: Write the dot product equation $\vec{a} \cdot \vec{c} = 3$
The dot product $ \vec{a} \cdot \vec{c} $ gives:
$$
(1,\,2,\,1)\,\cdot\,(x,\,y,\,z) = x + 2y + z = 3.
$$
This is our fourth equation.
Step 4: Solve the system of equations for $x$, $y$, and $z$
From $ x - z = -3 $, we get $ x = z - 3 $.
From $ y - 2x = 3 $, we have $ y = 3 + 2x $. Substituting $ x = z - 3 $ gives:
$$
y = 3 + 2(z - 3) = 3 + 2z - 6 = 2z - 3.
$$
Plug $ x = z - 3 $ and $ y = 2z - 3 $ into $ x + 2y + z = 3 $:
$$
(z - 3) + 2(2z - 3) + z = 3.
$$
Simplify:
$$
z - 3 + 4z - 6 + z = 3 \quad \Longrightarrow \quad 6z - 9 = 3 \quad \Longrightarrow \quad 6z = 12 \quad \Longrightarrow \quad z = 2.
$$
Then $ x = z - 3 = 2 - 3 = -1 $ and $ y = 2z - 3 = 4 - 3 = 1 $.
Therefore, $ \vec{c} = ( -1,\, 1,\, 2 ). $
Step 5: Calculate $\vec{c} \times \vec{b}$
$ \vec{c} = (-1,\, 1,\, 2) $ and $ \vec{b} = (3,\, -3,\, 3). $
$$
\vec{c} \times \vec{b}
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & 2 \\
3 & -3 & 3
\end{vmatrix}.
$$
Carefully computing each component:
$$
\vec{c} \times \vec{b}
= \Bigl(1 \cdot 3 - 2 \cdot (-3),\; -\bigl((-1)\cdot 3 - 2 \cdot 3\bigr),\;(-1)\cdot(-3) - 1\cdot 3\Bigr).
$$
So,
$$
\vec{c} \times \vec{b}
= (3 + 6,\, -(-3 - 6),\, 3 - 3) = (9,\, 9,\, 0).
$$
Step 6: Form the expression $(\vec{c} \times \vec{b}) - \vec{b} - \vec{c}$
First compute $(\vec{c} \times \vec{b}) - \vec{b}$:
$$
(9,\,9,\,0) - (3,\, -3,\, 3) = (9-3,\, 9-(-3),\, 0-3) = (6,\, 12,\, -3).
$$
Then subtract $\vec{c} = (-1,\,1,\,2)$:
$$
(6,\,12,\,-3) - (-1,\,1,\,2) = (6 - (-1),\, 12 - 1,\, -3 - 2) = (7,\, 11,\, -5).
$$
Step 7: Find $\vec{a} \cdot \bigl[(\vec{c} \times \vec{b}) - \vec{b} - \vec{c}\bigr]$
Recall $ \vec{a} = (1,\, 2,\, 1). $ Hence:
$$
\vec{a} \cdot (7,\, 11,\, -5)
= 1 \times 7 + 2 \times 11 + 1 \times (-5)
= 7 + 22 - 5
= 24.
$$
Final Answer
$ \vec{a} \cdot \bigl[(\vec{c} \times \vec{b}) - \vec{b} - \vec{c}\bigr] = 24.
