A plane is parallel to two lines whose direct | vedclass

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(C) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines: $\vec{n} = (2, 0, -2) 	imes (-2, 2, 0)$.

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$36$

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(C) The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines: $\vec{n} = (2, 0, -2) 	imes (-2, 2, 0)$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & -2 \ -2 & 2 & 0 \end{vmatrix} = \hat{i}(0 - (-4)) - \hat{j}(0 - 4) + \hat{k}(4 - 0) = 4\hat{i} + 4\hat{j} + 4\hat{k}$.We can take the normal vector as $\vec{n} = (1, 1, 1)$.The equation of the plane passing through $(2, 2, 2)$ is $1(x-2) + 1(y-2) + 1(z-2) = 0$,which simplifies to $x + y + z = 6$.The intercepts on the axes are $A(6, 0, 0)$,$B(0, 6, 0)$,and $C(0, 0, 6)$.The volume of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |x_{int} \cdot y_{int} \cdot z_{int}|$.$V = \frac{1}{6} |6 	imes 6 	imes 6| = \frac{216}{6} = 36 	ext{ cubic units}$.

$A$ plane is parallel to two lines whose direction ratios are $2, 0, -2$ and $-2, 2, 0$ and it contains the point $(2, 2, 2)$. If it cuts the coordinate axes at $A, B, C$,then the volume of the tetrahedron $OABC$ (in cubic units) is

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