A plane is parallel to two lines whose direct | vedclass

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(D) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines: $\vec{n} = (1, 0, -1) 	imes (-1, 1, 0)$.

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$\frac{9}{2}$

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(D) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines: $\vec{n} = (1, 0, -1) 	imes (-1, 1, 0)$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & -1 \ -1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(0 - 1) + \hat{k}(1 - 0) = \hat{i} + \hat{j} + \hat{k}$.Thus,the normal vector is $\langle 1, 1, 1 angle$.The equation of the plane passing through $(1, 1, 1)$ with normal $\langle 1, 1, 1 angle$ is $1(x-1) + 1(y-1) + 1(z-1) = 0$,which simplifies to $x + y + z = 3$.Dividing by $3$,we get $\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1$.The intercepts on the coordinate axes are $a = 3$,$b = 3$,and $c = 3$.The volume of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |abc| = \frac{1}{6} |3 	imes 3 	imes 3| = \frac{27}{6} = \frac{9}{2}$ cubic units.

$A$ plane is parallel to two lines whose direction ratios are $(1, 0, -1)$ and $(-1, 1, 0)$ and it contains the point $(1, 1, 1)$. If it cuts the coordinate axes at $A$,$B$,and $C$,then the volume of the tetrahedron $OABC$ is (in cubic units):

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