Let vec{a} = hat{i} + 2hat{j} + 3hat{k}, vec{ | vedclass

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(B) vector perpendicular to both $\vec{b}$ and $\vec{c}$ is given by $\vec{r} = \lambda(\vec{b} 	imes \vec{c})$.$\vec{b} 	imes \vec{c} = \begin{vmat

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

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(B) vector perpendicular to both $\vec{b}$ and $\vec{c}$ is given by $\vec{r} = \lambda(\vec{b} 	imes \vec{c})$.$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & 5 \ 1 & -4 & -2 \end{vmatrix} = \hat{i}(2 + 20) - \hat{j}(-6 - 5) + \hat{k}(-12 + 1) = 22\hat{i} + 11\hat{j} - 11\hat{k} = 11(2\hat{i} + \hat{j} - \hat{k})$.Thus, $\vec{r} = \lambda(11)(2\hat{i} + \hat{j} - \hat{k})$.Given $\vec{r} \cdot \vec{a} = 11$, we have $11\lambda(2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 11$.$11\lambda(2(1) + 1(2) - 1(3)) = 11 \implies 11\lambda(2 + 2 - 3) = 11 \implies 11\lambda(1) = 11 \implies \lambda = 1$.So, $\vec{r} = 11(2\hat{i} + \hat{j} - \hat{k})$.$A$ vector $\vec{p}$ is perpendicular to $\vec{r}$ if $\vec{r} \cdot \vec{p} = 0$, which implies $(2\hat{i} + \hat{j} - \hat{k}) \cdot \vec{p} = 0$.Checking option $B$: $(2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = 2(1) + 1(-1) - 1(1) = 2 - 1 - 1 = 0$.Thus, the vector $\hat{i} - \hat{j} + \hat{k}$ is perpendicular to $\vec{r}$.

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