If P→,q→ are r→ are three mutually perpendicular vectors of the same magnitude. If vector x→ satisfiesthe equation p→×((x→−q→)×p→)+q→×((x→−r→)×q→)+r→×((x→−p→)×r→)=0→, then x→ is given by

p→=ai^,q→=aj^,r→=ak^ Also, let x→=x1i^+y1j^+z1k^Given that x→ satisfies the equationp→×[(x→−q→)×p→]+q→×[(x→−r→)×q→]+[(x→−p→)×r→]=0…..(i) Now, p→×[(x→−q→)×p→]=p→×[x→×p→−q→×p→]=p→×(x→×p→)−p→×(q→×p→)=(p→⋅p→)x→−(p→x→)p→−(p→⋅p→)q→+(p→⋅q→)p→=a2x→−a2x1i^−a3j^+0 or 3a2x→−a2x→−a2(p→+q→+r→)=0→ or 2a2x→=(p→+q→+r→)a2 or x→=12(p→+q→+r→)

If P→,q→ are r→ are three mutually perpendicular vectors of the same magnitude. If vector x→ satisfiesthe equation p→×((x→−q→)×p→)+q→×((x→−r→)×q→)+r→×((x→−p→)×r→)=0→, then x→ is given by

p→=ai^,q→=aj^,r→=ak^ Also, let x→=x1i^+y1j^+z1k^Given that x→ satisfies the equationp→×[(x→−q→)×p→]+q→×[(x→−r→)×q→]+[(x→−p→)×r→]=0…..(i) Now, p→×[(x→−q→)×p→]=p→×[x→×p→−q→×p→]=p→×(x→×p→)−p→×(q→×p→)=(p→⋅p→)x→−(p→x→)p→−(p→⋅p→)q→+(p→⋅q→)p→=a2x→−a2x1i^−a3j^+0 or 3a2x→−a2x→−a2(p→+q→+r→)=0→ or 2a2x→=(p→+q→+r→)a2 or x→=12(p→+q→+r→)

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

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

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

$\frac{1}{3} (\vec{p} + \vec{q} + \vec{r})$

$\frac{1}{3} (2 \vec{p} + \vec{q} - \vec{r})$

answer is B.

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Detailed Solution

$\vec{p} = a \hat{i}, \vec{q} = a \hat{j}, \vec{r} = a \hat{k} Also, let \vec{x} = x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k}$

Given that $\vec{x}$ satisfies the equation

$\vec{p} \times [(\vec{x} - \vec{q}) \times \vec{p}] + \vec{q} \times [(\vec{x} - \vec{r}) \times \vec{q}] + [(\vec{x} - \vec{p}) \times \vec{r}] = 0 \dots . . (i)$