Key Concepts and Formulas

1. Vector Cross Product Properties:

   • Anti-commutativity: $\overrightarrow u \times \overrightarrow v = - (\overrightarrow v \times \overrightarrow u)$.
   • Condition for Parallel Vectors: If $\overrightarrow u \times \overrightarrow v = \overrightarrow 0$ and $\overrightarrow u, \overrightarrow v \neq \overrightarrow 0$, then $\overrightarrow u$ is parallel to $\overrightarrow v$ (i.e., $\overrightarrow u = k\overrightarrow v$ for some scalar $k$).
   • Distributive Property: $\overrightarrow u \times (\overrightarrow v + \overrightarrow w) = \overrightarrow u \times \overrightarrow v + \overrightarrow u \times \overrightarrow w$.

2. Vector Dot Product: For $\overrightarrow A = A_x \widehat i + A_y \widehat j + A_z \widehat k$ and $\overrightarrow B = B_x \widehat i + B_y \widehat j + B_z \widehat k$, $\overrightarrow A \cdot \overrightarrow B = A_x B_x + A_y B_y + A_z B_z$.

3. Magnitude of a Vector: For $\overrightarrow A = A_x \widehat i + A_y \widehat j + A_z \widehat k$, ${\left| {\overrightarrow A } \right|^2} = A_x^2 + A_y^2 + A_z^2$.

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Step-by-step Solution

Step 1: Simplify the given cross product equation to determine the relationship of $\overrightarrow r$ with $\overrightarrow a$ and $\overrightarrow b$. We are given the equation $\overrightarrow r \times \overrightarrow a = \overrightarrow b \times \overrightarrow r$. The goal here is to isolate $\overrightarrow r$ or find a relationship it satisfies. Rearrange the equation to bring all terms to one side: $\overrightarrow r \times \overrightarrow a - \overrightarrow b \times \overrightarrow r = \overrightarrow 0$ Using the anti-commutativity property of the cross product, $\overrightarrow b \times \overrightarrow r = -(\overrightarrow r \times \overrightarrow b)$: $\overrightarrow r \times \overrightarrow a + \overrightarrow r \times \overrightarrow b = \overrightarrow 0$ Now, using the distributive property of the cross product, we can factor out $\overrightarrow r$: $\overrightarrow r \times (\overrightarrow a + \overrightarrow b) = \overrightarrow 0$ This equation implies that $\overrightarrow r$ is parallel to the vector $(\overrightarrow a + \overrightarrow b)$, provided neither vector is the zero vector.

Let's compute $\overrightarrow a + \overrightarrow b$: Given $\overrightarrow a = \widehat i + 2\widehat j - 3\widehat k$ and $\overrightarrow b = 2\widehat i - 3\widehat j + 5\widehat k$. $\overrightarrow a + \overrightarrow b = (\widehat i + 2\widehat j - 3\widehat k) + (2\widehat i - 3\widehat j + 5\widehat k)$ $\overrightarrow a + \overrightarrow b = (1+2)\widehat i + (2-3)\widehat j + (-3+5)\widehat k$ $\overrightarrow a + \overrightarrow b = 3\widehat i - \widehat j + 2\widehat k$ Since $\overrightarrow a + \overrightarrow b \neq \overrightarrow 0$. If $\overrightarrow r = \overrightarrow 0$, then $\overrightarrow r \cdot (\alpha \widehat i + 2\widehat j + \widehat k) = 0$, which contradicts the given condition that it equals 3. Therefore, $\overrightarrow r \neq \overrightarrow 0$. Since $\overrightarrow r \times (\overrightarrow a + \overrightarrow b) = \overrightarrow 0$ and both vectors are non-zero, $\overrightarrow r$ must be parallel to $(\overrightarrow a + \overrightarrow b)$. Thus, we can write $\overrightarrow r$ as a scalar multiple of $(\overrightarrow a + \overrightarrow b)$: $\overrightarrow r = \lambda (\overrightarrow a + \overrightarrow b)$ where $\lambda$ is a non-zero scalar. Substituting the value of $\overrightarrow a + \overrightarrow b$: $\overrightarrow r = \lambda (3\widehat i - \widehat j + 2\widehat k)$ $\overrightarrow r = 3\lambda \widehat i - \lambda \widehat j + 2\lambda \widehat k$

Step 2: Use the given dot product conditions to form a system of equations involving $\lambda$ and $\alpha$. We are given two dot product conditions:

1. $\overrightarrow r \cdot (\alpha \widehat i + 2\widehat j + \widehat k) = 3$
2. $\overrightarrow r \cdot (2\widehat i + 5\widehat j - \alpha \widehat k) = -1$ Substituting the expression for $\overrightarrow r = 3\lambda \widehat i - \lambda \widehat j + 2\lambda \widehat k$ into these equations will give us scalar equations to solve for $\lambda$ and $\alpha$.

For the first condition: $(3\lambda \widehat i - \lambda \widehat j + 2\lambda \widehat k) \cdot (\alpha \widehat i + 2\widehat j + \widehat k) = 3$ Using the dot product formula: $(3\lambda)(\alpha) + (-\lambda)(2) + (2\lambda)(1) = 3$ $3\alpha\lambda - 2\lambda + 2\lambda = 3$ $3\alpha\lambda = 3$ Since $\lambda \neq 0$, we can divide by 3: $\alpha\lambda = 1 \quad \text{(Equation 1)}$

For the second condition: $(3\lambda \widehat i - \lambda \widehat j + 2\lambda \widehat k) \cdot (2\widehat i + 5\widehat j - \alpha \widehat k) = -1$ Using the dot product formula: $(3\lambda)(2) + (-\lambda)(5) + (2\lambda)(-\alpha) = -1$ $6\lambda - 5\lambda - 2\alpha\lambda = -1$ $\lambda - 2\alpha\lambda = -1 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations for $\lambda$ and $\alpha$. We have the system of equations:

1. $\alpha\lambda = 1$
2. $\lambda - 2\alpha\lambda = -1$ Substitute Equation 1 into Equation 2: $\lambda - 2(1) = -1$ $\lambda - 2 = -1$ Solving for $\lambda$: $\lambda = 1$ Now, substitute $\lambda = 1$ back into Equation 1 to find $\alpha$: $\alpha(1) = 1$ $\alpha = 1$

Step 4: Determine the vector $\overrightarrow r$ and its squared magnitude. Now that we have $\lambda = 1$, we can find the specific vector $\overrightarrow r$: $\overrightarrow r = \lambda (3\widehat i - \widehat j + 2\widehat k)$ $\overrightarrow r = 1 (3\widehat i - \widehat j + 2\widehat k)$ $\overrightarrow r = 3\widehat i - \widehat j + 2\widehat k$ Next, we calculate the squared magnitude of $\overrightarrow r$: ${\left| {\overrightarrow r } \right|^2} = (3)^2 + (-1)^2 + (2)^2$ ${\left| {\overrightarrow r } \right|^2} = 9 + 1 + 4$ ${\left| {\overrightarrow r } \right|^2} = 14$

Step 5: Calculate the final required value. The problem asks for the value of $\alpha + {\left| {\overrightarrow r } \right|^2}$. We found $\alpha = 1$ and ${\left| {\overrightarrow r } \right|^2} = 14$. $\alpha + {\left| {\overrightarrow r } \right|^2} = 1 + 14$ $\alpha + {\left| {\overrightarrow r } \right|^2} = 15$

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Common Mistakes & Tips

• Cross Product Simplification: Incorrectly applying cross product properties (like anti-commutativity) can lead to errors in determining the direction of $\overrightarrow r$. Always ensure the equation is set to zero before factoring.
• Algebraic Errors: Solving systems of equations, especially with multiple variables and signs, requires careful attention to detail. Double-check all arithmetic.
• Dot Product Calculation: Ensure accurate component-wise multiplication and summation when computing dot products.

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Summary The problem starts by simplifying the vector cross product equation $\overrightarrow r \times \overrightarrow a = \overrightarrow b \times \overrightarrow r$ to deduce that $\overrightarrow r$ is parallel to $(\overrightarrow a + \overrightarrow b)$. This allows us to express $\overrightarrow r$ as a scalar multiple, $\lambda(\overrightarrow a + \overrightarrow b)$. The two given dot product conditions then form a system of linear equations in $\lambda$ and $\alpha$. Solving this system yields the values of $\lambda$ and $\alpha$. Finally, with $\overrightarrow r$ determined and $\alpha$ found, we compute $\alpha + {\left| {\overrightarrow r } \right|^2}$.

The final answer is $\boxed{15}$.