Let overrightarrow{a}=hat{i}-2 hat{j}+3 hat{k | vedclass

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(A) Given vectors are $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$,$\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambd

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$0$

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(A) Given vectors are $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$,$\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$.Since $\overrightarrow{a}$ and $\overrightarrow{b}$ lie in a plane,their cross product $\overrightarrow{n} = \overrightarrow{a} 	imes \overrightarrow{b}$ is a vector perpendicular to that plane.If $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$,then $\overrightarrow{c}$ must be perpendicular to the normal vector $\overrightarrow{n}$.Therefore,the scalar triple product $(\overrightarrow{a} 	imes \overrightarrow{b}) \cdot \overrightarrow{c} = 0$.First,calculate $\overrightarrow{a} 	imes \overrightarrow{b}$:$\overrightarrow{a} 	imes \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2-9) - \hat{j}(-1-6) + \hat{k}(3+4) = -7\hat{i} + 7\hat{j} + 7\hat{k}$.Now,compute the dot product with $\overrightarrow{c}$:$(-7\hat{i} + 7\hat{j} + 7\hat{k}) \cdot (\lambda \hat{i} + \hat{j} + (2\lambda-1)\hat{k}) = 0$.$-7\lambda + 7(1) + 7(2\lambda-1) = 0$.$-7\lambda + 7 + 14\lambda - 7 = 0$.$7\lambda = 0 \Rightarrow \lambda = 0$.Thus,the value of $\lambda$ is $0$.

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