Let a line L passing through the point P(1, 1 | vedclass

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(B) The direction vectors of the two given lines are $\vec{d}_1 = \langle 4, 1, 1 \rangle$ and $\vec{d}_2 = \langle 1, 1, 0 \rangle$. Since line $L$ i

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$6$

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(B) The direction vectors of the two given lines are $\vec{d}_1 = \langle 4, 1, 1 angle$ and $\vec{d}_2 = \langle 1, 1, 0 angle$. Since line $L$ is perpendicular to both, its direction vector $\vec{d}_L = \vec{d}_1 	imes \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix} = \langle -1, 1, 3 angle$. The equation of line $L$ passing through $P(1, 1, 1)$ is $\vec{r}(t) = \langle 1-t, 1+t, 1+3t angle$. For point $Q$ on the $yz$-plane, $x = 0 \Rightarrow 1-t = 0 \Rightarrow t = 1$. Thus, $Q = (0, 2, 4)$. The line parallel to $L$ passing through $S(1, 0, -1)$ is $\vec{r}'(u) = \langle 1-u, u, -1+3u angle$. For point $R$ on the $yz$-plane, $x = 0 \Rightarrow 1-u = 0 \Rightarrow u = 1$. Thus, $R = (0, 1, 2)$. The parallelogram $PQRS$ has adjacent vectors $\vec{PQ} = Q - P = \langle -1, 1, 3 angle$ and $\vec{PS} = S - P = \langle 0, -1, -2 angle$. The area vector is $\vec{A} = \vec{PQ} 	imes \vec{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 3 \ 0 & -1 & -2 \end{vmatrix} = \langle 1, -2, 1 angle$. The area is $|\vec{A}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$. The square of the area is $(\sqrt{6})^2 = 6$.

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