1. Key Concepts and Formulas

• Vector Cross Product Properties:
  • $\vec{u} \times \vec{v} = \vec{0}$ if $\vec{u}$ and $\vec{v}$ are parallel (collinear).
  • $\vec{u} \times (\vec{v} - \vec{w}) = \vec{u} \times \vec{v} - \vec{u} \times \vec{w}$ (Distributivity).
  • $(\vec{u} - \vec{v}) \times \vec{w} = \vec{u} \times \vec{w} - \vec{v} \times \vec{w}$ (Distributivity).
• Vector Dot Product Properties:
  • $|\vec{u}|^2 = \vec{u} \cdot \vec{u}$.
  • $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$ (Distributivity).
  • $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$.
• Linear Independence: If $\vec{a}$ and $\vec{b}$ are non-collinear, then $p\vec{a} + q\vec{b} = \vec{0}$ implies $p=0$ and $q=0$.

2. Step-by-Step Solution

Step 1: Analyze the given cross product conditions to establish a relationship for $\vec{c}$. We are given $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$ and $\vec{a} \times \vec{b} = \vec{c} \times \vec{b}$.

From the first condition, $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$: $\vec{a} \times \vec{c} - \vec{a} \times \vec{b} = \vec{0}$ Using the distributive property of the cross product: $\vec{a} \times (\vec{c} - \vec{b}) = \vec{0}$ Since $\vec{a}$ is non-zero, this implies that $(\vec{c} - \vec{b})$ is parallel to $\vec{a}$. Thus, we can write: $\vec{c} - \vec{b} = k_1 \vec{a} \quad \text{for some scalar } k_1$ $\vec{c} = \vec{b} + k_1 \vec{a} \quad (*)$

From the second condition, $\vec{a} \times \vec{b} = \vec{c} \times \vec{b}$: $\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = \vec{0}$ Rearranging the terms using anticommutativity ($\vec{c} \times \vec{b} = - \vec{b} \times \vec{c}$): $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$ This doesn't directly give us a useful form for $\vec{c}$ without further manipulation. Let's use the original form and rearrange: $\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = \vec{0}$ Using the distributive property: $(\vec{a} - \vec{c}) \times \vec{b} = \vec{0}$ Since $\vec{b}$ is non-zero, this implies that $(\vec{a} - \vec{c})$ is parallel to $\vec{b}$. Thus, we can write: $\vec{a} - \vec{c} = k_2 \vec{b} \quad \text{for some scalar } k_2$ $\vec{c} = \vec{a} - k_2 \vec{b} \quad (**)$

Step 2: Equate the expressions for $\vec{c}$ to find the relationship between $k_1$ and $k_2$. From $(*)$ and $(**)$, we have: $\vec{b} + k_1 \vec{a} = \vec{a} - k_2 \vec{b}$ Rearranging the terms to group $\vec{a}$ and $\vec{b}$: $k_1 \vec{a} - \vec{a} + \vec{b} + k_2 \vec{b} = \vec{0}$ $(k_1 - 1)\vec{a} + (1 + k_2)\vec{b} = \vec{0}$

Step 3: Determine if $\vec{a}$ and $\vec{b}$ are collinear and use this to find $k_1$ and $k_2$. Given $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + \hat{k}$. For $\vec{a}$ and $\vec{b}$ to be collinear, there must exist a scalar $m$ such that $\vec{b} = m\vec{a}$. Comparing components: $3 = 2m \implies m = 3/2$ $-5 = -m \implies m = 5$ $1 = 3m \implies m = 1/3$ Since the values of $m$ are inconsistent, $\vec{a}$ and $\vec{b}$ are not collinear (i.e., they are linearly independent).

Since $\vec{a}$ and $\vec{b}$ are linearly independent, for the equation $(k_1 - 1)\vec{a} + (1 + k_2)\vec{b} = \vec{0}$ to hold, the coefficients must be zero: $k_1 - 1 = 0 \implies k_1 = 1$ $1 + k_2 = 0 \implies k_2 = -1$

Step 4: Find the explicit expression for $\vec{c}$. Substitute $k_1 = 1$ into the expression $\vec{c} = \vec{b} + k_1 \vec{a}$: $\vec{c} = \vec{b} + 1 \cdot \vec{a}$ $\vec{c} = \vec{a} + \vec{b}$ This shows that $\vec{c}$ is uniquely determined as the sum of $\vec{a}$ and $\vec{b}$.

Step 5: Calculate the components of $\vec{c}$ and its squared magnitude. Given $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + \hat{k}$. $\vec{c} = \vec{a} + \vec{b} = (2\hat{i} - \hat{j} + 3\hat{k}) + (3\hat{i} - 5\hat{j} + \hat{k})$ $\vec{c} = (2+3)\hat{i} + (-1-5)\hat{j} + (3+1)\hat{k}$ $\vec{c} = 5\hat{i} - 6\hat{j} + 4\hat{k}$ Now, we calculate $|\vec{c}|^2$: $|\vec{c}|^2 = (5)^2 + (-6)^2 + (4)^2$ $|\vec{c}|^2 = 25 + 36 + 16$ $|\vec{c}|^2 = 77$

Step 6: Verify the dot product condition using the determined $\vec{c}$. The condition is $(\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168$. Substitute $\vec{c} = \vec{a} + \vec{b}$: $(\vec{a} + (\vec{a} + \vec{b})) \cdot (\vec{b} + (\vec{a} + \vec{b})) = 168$ $(2\vec{a} + \vec{b}) \cdot (\vec{a} + 2\vec{b}) = 168$ Expand the dot product: $2\vec{a} \cdot \vec{a} + 4\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + 2\vec{b} \cdot \vec{b} = 168$ $2|\vec{a}|^2 + 5(\vec{a} \cdot \vec{b}) + 2|\vec{b}|^2 = 168$ Calculate the required terms: $|\vec{a}|^2 = 2^2 + (-1)^2 + 3^2 = 4 + 1 + 9 = 14$. $|\vec{b}|^2 = 3^2 + (-5)^2 + 1^2 = 9 + 25 + 1 = 35$. $\vec{a} \cdot \vec{b} = (2)(3) + (-1)(-5) + (3)(1) = 6 + 5 + 3 = 14$. Substitute these values: $2(14) + 5(14) + 2(35) = 28 + 70 + 70 = 168$ The condition is satisfied, confirming our expression for $\vec{c}$.

3. Common Mistakes & Tips

• Overlooking unique determination of $\vec{c}$: A common mistake is to conclude from $\vec{a} \times (\vec{c} - \vec{b}) = \vec{0}$ that $\vec{c} - \vec{b} = k\vec{a}$ and from $(\vec{a} - \vec{c}) \times \vec{b} = \vec{0}$ that $\vec{a} - \vec{c} = m\vec{b}$. However, simply combining these without considering linear independence of $\vec{a}$ and $\vec{b}$ can lead to errors. Using both conditions simultaneously and equating the forms of $\vec{c}$ is crucial.
• Assuming $\vec{c}$ is parallel to $\vec{a} + \vec{b}$: If one only considered $(\vec{a}+\vec{b}) \times \vec{c} = \vec{0}$ (which can be derived by adding cross products), it would imply $\vec{c} = \lambda(\vec{a}+\vec{b})$, where $\lambda$ could be any scalar. The given conditions uniquely fix $\lambda=1$.
• Algebraic errors in dot product expansion: Carefully expanding $(\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c})$ requires attention to detail to avoid sign errors or missed terms.

4. Summary

The problem provides conditions involving vector cross products and a dot product. By analyzing the cross product conditions $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$ and $\vec{a} \times \vec{b} = \vec{c} \times \vec{b}$, we deduced that $\vec{c} - \vec{b}$ must be parallel to $\vec{a}$ and $\vec{a} - \vec{c}$ must be parallel to $\vec{b}$. By expressing $\vec{c}$ in two ways and using the linear independence of $\vec{a}$ and $\vec{b}$, we uniquely determined that $\vec{c} = \vec{a} + \vec{b}$. This result was verified using the given dot product condition. Finally, the magnitude squared of $\vec{c}$ was calculated. Since $\vec{c}$ is uniquely determined, its magnitude squared has a single value, which is thus the maximum value.

The maximum value of $|\vec{c}|^2$ is $77$.

5. Final Answer

The final answer is $\boxed{77}$.