Let L_1: overrightarrow{r}=(hat{i}-hat{j}+2 h | vedclass

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(A) Given $L_1$ is perpendicular to $L_2$. The direction vectors are $\vec{v_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{v_2} = 3\hat{i} + \hat{j} +

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$(-1, 7, 4)$

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(A) Given $L_1$ is perpendicular to $L_2$. The direction vectors are $\vec{v_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{v_2} = 3\hat{i} + \hat{j} + p\hat{k}$.Since $L_1 \perp L_2$,their dot product is zero: $(1)(3) + (-1)(1) + (2)(p) = 0$.$3 - 1 + 2p = 0 \implies 2p = -2 \implies p = -1$.Line $L_3$ is perpendicular to both $L_1$ and $L_2$,so its direction vector $\vec{v_3}$ is parallel to $\vec{v_1} 	imes \vec{v_2}$.$\vec{v_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 3 & 1 & -1 \end{vmatrix} = \hat{i}(1 - 2) - \hat{j}(-1 - 6) + \hat{k}(1 + 3) = -\hat{i} + 7\hat{j} + 4\hat{k}$.Thus,the equation of $L_3$ is $\overrightarrow{r} = \delta(-\hat{i} + 7\hat{j} + 4\hat{k})$.Any point on $L_3$ is of the form $(-\delta, 7\delta, 4\delta)$.For $\delta = 1$,the point is $(-1, 7, 4)$.

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