Step 1: Understanding the Concept:
Two vectors are perpendicular if their dot product is zero. We will use the properties of scalar triple products, specifically that \( (\vec{a} \times \vec{b}) \cdot \vec{a} = 0 \), and only cross products with the "missing" vector in the dot product return a non-zero scalar triple product \( [\vec{a} \vec{b} \vec{c}] \).
Step 2: Key Formula or Approach:
1. \( \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \).
2. Solve for \( x \) and \( y \) given non-coplanar vectors (\( [\vec{a} \vec{b} \vec{c}] \neq 0 \)).
Step 3: Detailed Explanation:
Expand the dot product:
\( \vec{r} \cdot \vec{a} + \vec{r} \cdot \vec{b} + \vec{r} \cdot \vec{c} = 0 \).
\( \vec{r} \cdot \vec{a} = \cos y(\vec{b} \times \vec{c} \cdot \vec{a}) = \cos y [\vec{a} \vec{b} \vec{c}] \).
\( \vec{r} \cdot \vec{b} = 2(\vec{c} \times \vec{a} \cdot \vec{b}) = 2 [\vec{a} \vec{b} \vec{c}] \).
\( \vec{r} \cdot \vec{c} = \sin x(\vec{a} \times \vec{b} \cdot \vec{c}) = \sin x [\vec{a} \vec{b} \vec{c}] \).
Summing these up:
\[ (\cos y + 2 + \sin x) [\vec{a} \vec{b} \vec{c}] = 0 \]
Since vectors are non-coplanar, \( [\vec{a} \vec{b} \vec{c}] \neq 0 \).
\[ \sin x + \cos y = -2 \]
The range of both \( \sin \) and \( \cos \) is \( [-1, 1] \). For the sum to be -2, both must be at their minimum value:
\( \sin x = -1 \implies x = 2n\pi - \pi/2 \).
\( \cos y = -1 \implies y = (2m+1)\pi \).
Test for smallest values:
If \( n=0 \), \( x = -\pi/2 \). If \( m=0 \), \( y = \pi \).
\( x^2 + y^2 = \pi^2/4 + \pi^2 = 5\pi^2 / 4 \). (A) is correct.
If \( n=0 \), \( x = -\pi/2 \). If \( m=1, y = 3\pi \).
\( x^2 + y^2 = \pi^2/4 + 9\pi^2 = 37\pi^2 / 4 \). (C) is correct.
Step 4: Final Answer:
The possible values are \( 5\pi^2/4 \) and \( 37\pi^2/4 \).