A plane P is parallel to two lines whose dire | vedclass

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(B) The normal vector $\vec{n}$ to the plane $P$ is the cross product of the direction vectors of the two lines:$\vec{n} = (-2, 1, -3) 	imes (-1, 2,

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$(24, -13)$

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(B) The normal vector $\vec{n}$ to the plane $P$ is the cross product of the direction vectors of the two lines:$\vec{n} = (-2, 1, -3) 	imes (-1, 2, -2) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 1 & -3 \ -1 & 2 & -2 \end{vmatrix} = \hat{i}(-2 + 6) - \hat{j}(4 - 3) + \hat{k}(-4 + 1) = 4\hat{i} - \hat{j} - 3\hat{k}$.The equation of the plane passing through $(2, 2, -2)$ with normal vector $(4, -1, -3)$ is:$4(x - 2) - 1(y - 2) - 3(z + 2) = 0$$4x - 8 - y + 2 - 3z - 6 = 0$$4x - y - 3z = 12$.To find the intercepts,divide by $12$:$\frac{x}{3} + \frac{y}{-12} + \frac{z}{-4} = 1$.Thus,$\alpha = 3, \beta = -12, \gamma = -4$.Calculating $p = \alpha + \beta + \gamma = 3 - 12 - 4 = -13$.The volume $V$ of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |\alpha \beta \gamma| = \frac{1}{6} |3 	imes (-12) 	imes (-4)| = \frac{1}{6} |144| = 24$.Therefore,the ordered pair $(V, p) = (24, -13)$.

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