Let a→=i^+j^;b→=2i^−k^ then vector r→ satisfying the equations r→×a→=b→×a→ and r→×b→=a→×b→ is

r→×a→=b→×a→ or (r→−b→)×a→=0r→×b→=a→×b→ or (r→−a→)×b→=0 If r→=xi^+yj^+zk^, then i^j^k^x−2yz+1110=0 and i^j^k^x−1y−1z20−1=0⇒ z+1=0,x−y=2 and y−1=0,x−1+2z=0⇒ x=3,y=1,z=−1

Let a→=i^+j^;b→=2i^−k^ then vector r→ satisfying the equations r→×a→=b→×a→ and r→×b→=a→×b→ is

r→×a→=b→×a→ or (r→−b→)×a→=0r→×b→=a→×b→ or (r→−a→)×b→=0 If r→=xi^+yj^+zk^, then i^j^k^x−2yz+1110=0 and i^j^k^x−1y−1z20−1=0⇒ z+1=0,x−y=2 and y−1=0,x−1+2z=0⇒ x=3,y=1,z=−1

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Let $\vec{a} = \hat{i} + \hat{j}; \vec{b} = 2 \hat{i} - \hat{k}$ then vector $\vec{r}$ satisfying the equations $\vec{r} \times \vec{a} = \vec{b} \times \vec{a} and \vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is

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$3 \hat{i} - \hat{j} + \hat{k}$

$\hat{i} - \hat{j} + \hat{k}$

$\hat{i} - \hat{j} - \hat{k}$

$3 \hat{i} + \hat{j} - \hat{k}$

answer is C.

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Detailed Solution

$\begin{aligned} \vec{r} \times \vec{a} = \vec{b} \times \vec{a} or (\vec{r} - \vec{b}) \times \vec{a} = 0 \\ \vec{r} \times \vec{b} = \vec{a} \times \vec{b} or (\vec{r} - \vec{a}) \times \vec{b} = 0 \end{aligned}$

$If \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}, then$

$\begin{array}{l} |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ x - 2 & y & z + 1 \\ 1 & 1 & 0 \end{array}| = 0 and |\begin{array}{clc} \hat{i} & \hat{j} & \hat{k} \\ x - 1 & y - 1 & z \\ 2 & 0 & - 1 \end{array}| = 0 \\ \Rightarrow z + 1 = 0, x - y = 2 \\ and y - 1 = 0, x - 1 + 2 z = 0 \\ \Rightarrow x = 3, y = 1, z = - 1 \end{array}$