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(B) Given that $p, q, r$ are mutually perpendicular vectors of the same magnitude,let $|p| = |q| = |r| = c$. Thus,$p \cdot q = q \cdot r = r \cdot p =

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$\frac{1}{2}(p + q + r)$

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(B) Given that $p, q, r$ are mutually perpendicular vectors of the same magnitude,let $|p| = |q| = |r| = c$. Thus,$p \cdot q = q \cdot r = r \cdot p = 0$ and $p \cdot p = q \cdot q = r \cdot r = c^2$. Using the vector triple product identity $a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$,we expand each term: $p 	imes \{(x - q) 	imes p\} = (p \cdot p)(x - q) - (p \cdot (x - q))p = c^2(x - q) - (p \cdot x)p$. Similarly,$q 	imes \{(x - r) 	imes q\} = c^2(x - r) - (q \cdot x)q$ and $r 	imes \{(x - p) 	imes r\} = c^2(x - p) - (r \cdot x)r$. Summing these,we get: $c^2(x - q + x - r + x - p) - [(p \cdot x)p + (q \cdot x)q + (r \cdot x)r] = 0$. $c^2(3x - (p + q + r)) - [(p \cdot x)p + (q \cdot x)q + (r \cdot x)r] = 0$. If we substitute $x = \frac{1}{2}(p + q + r)$,then $p \cdot x = \frac{1}{2}c^2$,$q \cdot x = \frac{1}{2}c^2$,and $r \cdot x = \frac{1}{2}c^2$. Substituting these into the equation: $c^2(3(\frac{1}{2}(p + q + r)) - (p + q + r)) - [\frac{1}{2}c^2 p + \frac{1}{2}c^2 q + \frac{1}{2}c^2 r] = 0$. $c^2(\frac{3}{2}(p + q + r) - (p + q + r)) - \frac{1}{2}c^2(p + q + r) = 0$. $c^2(\frac{1}{2}(p + q + r)) - \frac{1}{2}c^2(p + q + r) = 0$. This confirms $x = \frac{1}{2}(p + q + r)$ is the solution.

Let $p, q, r$ be three mutually perpendicular vectors of the same magnitude. If a vector $x$ satisfies the equation $p \times \{(x - q) \times p\} + q \times \{(x - r) \times q\} + r \times \{(x - p) \times r\} = 0$,then $x$ is given by

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