1. Key Concepts and Formulas

• Vector Perpendicularity: A vector $\overrightarrow r$ is perpendicular to vectors $\overrightarrow A$ and $\overrightarrow B$ if their dot products are zero: $\overrightarrow r \cdot \overrightarrow A = 0$ and $\overrightarrow r \cdot \overrightarrow B = 0$.
• Cross Product: The cross product of two vectors $\overrightarrow A$ and $\overrightarrow B$, denoted by $\overrightarrow A \times \overrightarrow B$, results in a vector that is perpendicular to both $\overrightarrow A$ and $\overrightarrow B$. If $\overrightarrow r$ is perpendicular to both $\overrightarrow A$ and $\overrightarrow B$, then $\overrightarrow r$ must be parallel to $\overrightarrow A \times \overrightarrow B$. Thus, $\overrightarrow r = k(\overrightarrow A \times \overrightarrow B)$ for some scalar $k$.
• Magnitude of a Vector: For a vector $\overrightarrow V = a\widehat i + b\widehat j + c\widehat k$, its magnitude is $|\overrightarrow V| = \sqrt{a^2 + b^2 + c^2}$.
• Scalar Multiplication of Vectors: If $\overrightarrow V = a\widehat i + b\widehat j + c\widehat k$ and $k$ is a scalar, then $k\overrightarrow V = (ka)\widehat i + (kb)\widehat j + (kc)\widehat k$.

2. Step-by-Step Solution

Step 1: Calculate the vectors $\overrightarrow p + \overrightarrow q$ and $\overrightarrow p - \overrightarrow q$. We are given $\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$ and $\overrightarrow q = \widehat i + 2\widehat j + \widehat k$. Let $\overrightarrow A = \overrightarrow p + \overrightarrow q$ and $\overrightarrow B = \overrightarrow p - \overrightarrow q$. $\overrightarrow A = (2\widehat i + 3\widehat j + \widehat k) + (\widehat i + 2\widehat j + \widehat k) = (2+1)\widehat i + (3+2)\widehat j + (1+1)\widehat k = 3\widehat i + 5\widehat j + 2\widehat k$ $\overrightarrow B = (2\widehat i + 3\widehat j + \widehat k) - (\widehat i + 2\widehat j + \widehat k) = (2-1)\widehat i + (3-2)\widehat j + (1-1)\widehat k = 1\widehat i + 1\widehat j + 0\widehat k$

Step 2: Find a vector perpendicular to $\overrightarrow A$ and $\overrightarrow B$ using the cross product. Since $\overrightarrow r$ is perpendicular to both $\overrightarrow A$ and $\overrightarrow B$, it must be parallel to their cross product, $\overrightarrow A \times \overrightarrow B$. $\overrightarrow A \times \overrightarrow B = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 3 & 5 & 2 \\ 1 & 1 & 0 \end{vmatrix}$ $= \widehat i (5 \cdot 0 - 2 \cdot 1) - \widehat j (3 \cdot 0 - 2 \cdot 1) + \widehat k (3 \cdot 1 - 5 \cdot 1)$ $= \widehat i (0 - 2) - \widehat j (0 - 2) + \widehat k (3 - 5)$ $= -2\widehat i + 2\widehat j - 2\widehat k$ So, $\overrightarrow r$ is parallel to $-2\widehat i + 2\widehat j - 2\widehat k$. We can write $\overrightarrow r = k(-2\widehat i + 2\widehat j - 2\widehat k)$ for some scalar $k$.

Step 3: Use the magnitude of $\overrightarrow r$ to find the scalar $k$. We are given $|\overrightarrow r| = \sqrt{3}$. The magnitude of the direction vector $-2\widehat i + 2\widehat j - 2\widehat k$ is: $|-2\widehat i + 2\widehat j - 2\widehat k| = \sqrt{(-2)^2 + (2)^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$ Now, we use the relation $|\overrightarrow r| = |k| |-2\widehat i + 2\widehat j - 2\widehat k|$: $\sqrt{3} = |k| (2\sqrt{3})$ Solving for $|k|$: $|k| = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}$ This means $k = \frac{1}{2}$ or $k = -\frac{1}{2}$.

Step 4: Determine the components $\alpha, \beta, \gamma$ of $\overrightarrow r$. Let's choose $k = \frac{1}{2}$. Then: $\overrightarrow r = \frac{1}{2}(-2\widehat i + 2\widehat j - 2\widehat k) = -1\widehat i + 1\widehat j - 1\widehat k$ Comparing this with $\overrightarrow r = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$, we have: $\alpha = -1, \quad \beta = 1, \quad \gamma = -1$ If we had chosen $k = -\frac{1}{2}$, we would get $\alpha = 1, \beta = -1, \gamma = 1$. The absolute values remain the same.

Step 5: Calculate $|\alpha| + |\beta| + |\gamma|$. Using the values from $k=\frac{1}{2}$: $|\alpha| + |\beta| + |\gamma| = |-1| + |1| + |-1| = 1 + 1 + 1 = 3$ Using the values from $k=-\frac{1}{2}$: $|\alpha| + |\beta| + |\gamma| = |1| + |-1| + |1| = 1 + 1 + 1 = 3$ In both cases, the result is 3.

3. Common Mistakes & Tips

• Sign Errors in Cross Product: Be meticulous when calculating the determinant for the cross product, especially with the middle term (the $\widehat j$ component) which has a negative sign.
• Forgetting the Scalar Factor: Remember that $\overrightarrow r$ is parallel to the cross product, not necessarily equal to it. The scalar factor $k$ is determined by the magnitude condition.
• Absolute Values: The final calculation requires the sum of the absolute values of the components of $\overrightarrow r$, so ensure you take the absolute value of each component before summing.

4. Summary

The problem requires finding a vector $\overrightarrow r$ that is perpendicular to two given vector combinations, $(\overrightarrow p + \overrightarrow q)$ and $(\overrightarrow p - \overrightarrow q)$, and has a specific magnitude. We first computed these two vectors. A vector perpendicular to two other vectors is found by taking their cross product. We calculated the cross product of $(\overrightarrow p + \overrightarrow q)$ and $(\overrightarrow p - \overrightarrow q)$ to get the direction of $\overrightarrow r$. Then, using the given magnitude of $\overrightarrow r$, we determined the scalar multiplier needed to scale the direction vector to the correct length. Finally, we extracted the components of $\overrightarrow r$ and calculated the sum of their absolute values.

The final answer is $\boxed{3}$.