Let $\overrightarrow{\mathrm{a}}=2\hat{\imath}-\hat{\jmath}+3\hat{k}$, $\overrightarrow{\mathrm{b}}=3\hat{\imath}-5\hat{\jmath}+\hat{k}$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$ and $(\vec{a}+\vec{c}) \cdot (\vec{b}+\vec{c})=168$. Then the maximum value of $|\vec{c}|^2$ is:

Step 1: Simplify the cross product equation

The given cross product equation is $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$. We know that the cross product is anti-commutative, meaning $\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}} = -(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})$. Substituting this into the equation and rearranging terms, we get $(\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}} = \overrightarrow{0}$. This implies that $\overrightarrow{\mathrm{c}}$ is parallel to $(\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}})$.

Step 2: Calculate $\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}$ and express $\overrightarrow{\mathrm{c}}$

First, we calculate the sum of vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$. Since $\overrightarrow{\mathrm{c}}$ is parallel to $(\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}})$, we can express $\overrightarrow{\mathrm{c}}$ as a scalar multiple of $(\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}})$, where $\lambda$ is a scalar.

Step 3: Expand the dot product equation

We are given the dot product equation $(\vec{a}+\vec{c}) \cdot (\vec{b}+\vec{c})=168$. Expanding this equation, we get $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 168$. We can rewrite the middle terms as $(\vec{a} + \vec{b}) \cdot \vec{c}$ and $\vec{c} \cdot \vec{c}$ is simply $|\vec{c}|^2$.

Step 4: Calculate dot products and magnitudes

We calculate the dot product $\vec{a} \cdot \vec{b}$. Then, we use the fact that $\vec{c} = \lambda (\vec{a} + \vec{b})$ to simplify $(\vec{a} + \vec{b}) \cdot \vec{c}$ to $\lambda |\vec{a} + \vec{b}|^2$. We also calculate the magnitude squared of $(\vec{a} + \vec{b})$ and express $|\vec{c}|^2$ in terms of $\lambda$.

Step 5: Substitute into the expanded dot product equation

Substitute the calculated values into the expanded dot product equation. This results in a quadratic equation in terms of $\lambda$. Solving this quadratic equation by factoring gives us two possible values for $\lambda$: $1$ and $-2$.

Step 6: Find the maximum value of $|\vec{c}|^2$

We use the expression for $|\vec{c}|^2 = 77\lambda^2$ and substitute the two values of $\lambda$ we found. For $\lambda = 1$, $|\vec{c}|^2 = 77$. For $\lambda = -2$, $|\vec{c}|^2 = 308$. The maximum value is the larger of these two.

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