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Step-by-Step Solution
Step 1: Identify the Given Vectors
We have
$$
\overrightarrow{a} = \hat{i} + \hat{j} + 2\hat{k},
\quad
\overrightarrow{b} = -\hat{i} + 2\hat{j} + 3\hat{k}.
$$
Step 2: Compute $\overrightarrow{a} + \overrightarrow{b}$
Add the vectors component-wise:
$$
\overrightarrow{a} + \overrightarrow{b}
\;=\; (\hat{i} - \hat{i}) + (\,\hat{j} + 2\hat{j}\,) + (\,2\hat{k} + 3\hat{k}\,)
\;=\; 0\hat{i} + 3\hat{j} + 5\hat{k}.
$$
Step 3: Simplify the Inner Vector Expression
We want to evaluate:
$$
\bigl(\overrightarrow{a} \times \bigl(\,(\overrightarrow{a} - \overrightarrow{b}) \times \overrightarrow{b}\bigr)\bigr)
\,\times\, \overrightarrow{b}.
$$
Notice that
$$
(\overrightarrow{a} - \overrightarrow{b}) \times \overrightarrow{b}
\;=\; (\overrightarrow{a} \times \overrightarrow{b}) - (\overrightarrow{b} \times \overrightarrow{b}).
$$
But $\overrightarrow{b}\times\overrightarrow{b} = \overrightarrow{0}$, so this reduces to
$$
(\overrightarrow{a} - \overrightarrow{b}) \times \overrightarrow{b}
\;=\; \overrightarrow{a} \times \overrightarrow{b}.
$$
Hence the expression becomes
$$
(\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{b})) \,\times\, \overrightarrow{b}.
$$
Step 4: Use the Vector Triple Product Identity
The identity for $\overrightarrow{x} \times (\overrightarrow{y} \times \overrightarrow{z})$ can be written as
$$
\overrightarrow{x} \times (\overrightarrow{y} \times \overrightarrow{z})
\;=\; (\overrightarrow{x} \cdot \overrightarrow{z})\,\overrightarrow{y}
\;-\; (\overrightarrow{x} \cdot \overrightarrow{y})\,\overrightarrow{z}.
$$
In particular,
$$
\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{b})
\;=\; (\overrightarrow{a}\cdot \overrightarrow{b})\,\overrightarrow{a}
\,-\, (\overrightarrow{a}\cdot \overrightarrow{a})\,\overrightarrow{b}.
$$
Then we take that result and cross it with $\overrightarrow{b}$:
$$
\bigl(\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{b})\bigr)
\times \overrightarrow{b}
\;=\;
\bigl((\overrightarrow{a}\cdot \overrightarrow{b})\,\overrightarrow{a} - (\overrightarrow{a}\cdot \overrightarrow{a})\,\overrightarrow{b}\bigr)
\times \overrightarrow{b}.
$$
Step 5: Observe the Term $(\overrightarrow{b} \times \overrightarrow{b})$
Notice that when we expand
$$
(\overrightarrow{a}\cdot \overrightarrow{b})\,\overrightarrow{a} \times \overrightarrow{b}
\;-\;
(\overrightarrow{a}\cdot \overrightarrow{a})\,(\overrightarrow{b} \times \overrightarrow{b}),
$$
the second part vanishes because $\overrightarrow{b} \times \overrightarrow{b} = \overrightarrow{0}.$ Hence,
$$
\bigl(\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{b})\bigr)
\times \overrightarrow{b}
\;=\;
(\overrightarrow{a}\cdot \overrightarrow{b})\,(\overrightarrow{a} \times \overrightarrow{b}).
$$
Step 6: Calculate $\overrightarrow{a}\cdot \overrightarrow{b}$ and $\overrightarrow{a}\times \overrightarrow{b}$
(a) Dot product:
$$
\overrightarrow{a}\cdot \overrightarrow{b}
= (1)(-1) + (1)(2) + (2)(3)
= -1 + 2 + 6
= 7.
$$
(b) Cross product: Using the determinant form,
$$
\overrightarrow{a} \times \overrightarrow{b}
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 2 \\
-1 & 2 & 3
\end{vmatrix}.
$$
Expand this determinant:
$$
\overrightarrow{a} \times \overrightarrow{b}
= \hat{i}\,(1\cdot 3 - 2\cdot 2)
\;-\;\hat{j}\,(1\cdot 3 - 2\cdot(-1))
\;+\;\hat{k}\,(1\cdot 2 - 1\cdot(-1))
$$
$$
= \hat{i}\,(3 - 4)
\;-\;\hat{j}\,(3 + 2)
\;+\;\hat{k}\,(2 + 1)
$$
$$
= -\hat{i} - 5\hat{j} + 3\hat{k}.
$$
Step 7: Combine the Results
Since
$$
\bigl(\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{b})\bigr)
\times \overrightarrow{b}
\;=\; (\overrightarrow{a}\cdot \overrightarrow{b})\,(\overrightarrow{a} \times \overrightarrow{b}),
$$
we have
$$
(\overrightarrow{a}\cdot \overrightarrow{b})\,(\overrightarrow{a} \times \overrightarrow{b})
\;=\;
7\bigl(-\hat{i} - 5\hat{j} + 3\hat{k}\bigr).
$$
Step 8: Final Cross Product with $(\overrightarrow{a} + \overrightarrow{b})$
We now compute
$$
(\overrightarrow{a} + \overrightarrow{b})
\times
\Bigl[(\overrightarrow{a} \times ((\overrightarrow{a}-\overrightarrow{b})\times\overrightarrow{b})) \times \overrightarrow{b}\Bigr]
\;=\;
(\overrightarrow{a} + \overrightarrow{b})
\times
\Bigl[\,7(-\hat{i} - 5\hat{j} + 3\hat{k})\Bigr].
$$
Recall from Step 2 that
$$
\overrightarrow{a} + \overrightarrow{b} = 0\hat{i} + 3\hat{j} + 5\hat{k}.
$$
Hence the final cross product becomes
$$
7 \Bigl( (\,0\hat{i} + 3\hat{j} + 5\hat{k}\,)
\times
(-\hat{i} - 5\hat{j} + 3\hat{k})\Bigr).
$$
Step 9: Expand the Determinant
Inside, we have
$$
(\,0\hat{i} + 3\hat{j} + 5\hat{k}\,)
\times
(-\hat{i} - 5\hat{j} + 3\hat{k}).
$$
Using the determinant form:
$$
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
0 & 3 & 5 \\
-1 & -5 & 3
\end{vmatrix}.
$$
Expand:
$\hat{i}\,(3\cdot 3 - 5\cdot(-5))
= \hat{i}\,(9 + 25)
= 34\hat{i}.$
$-\hat{j}\,(0\cdot 3 - 5\cdot(-1))
= -\hat{j}\,(0 + 5)
= -5\hat{j}.$
$\hat{k}\,(0\cdot(-5) - 3\cdot(-1))
= \hat{k}\,(0 + 3)
= 3\hat{k}.$
Summing gives
$$
34\hat{i}
- 5\hat{j}
+ 3\hat{k}.
$$
Step 10: Multiply by 7
Therefore,
$$
7 \Bigl(34\hat{i} - 5\hat{j} + 3\hat{k}\Bigr)
\;=\;
7(34)\hat{i} \;-\; 7(5)\hat{j} \;+\; 7(3)\hat{k}
\;=\;
7(34)\hat{i} - 35\hat{j} + 21\hat{k}.
$$
In compact form, the final answer is:
$$
\boxed{7\bigl(34\hat{i} \;-\; 5\hat{j} \;+\; 3\hat{k}\bigr).}
$$
Hence, the required vector product is $7(34\hat{i} - 5\hat{j} + 3\hat{k})$.
