Key Concepts and Formulas

1. Normal Vector to a Plane: If a plane is parallel to two lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$, its normal vector $\vec{n}$ is perpendicular to both. This means $\vec{n}$ can be found by taking their cross product: $\vec{n} = \vec{d_1} \times \vec{d_2}$.
2. Equation of a Plane: The equation of a plane passing through a point $(x_0, y_0, z_0)$ with a normal vector $\vec{n} = (A, B, C)$ is given by $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$.
3. Intercept Form of a Plane: A plane intersecting the $x, y, z$ axes at $\alpha, \beta, \gamma$ respectively has the equation $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$.
4. Volume of a Tetrahedron OABC: For a tetrahedron with vertices at the origin $O(0,0,0)$ and points $A(\alpha, 0, 0)$, $B(0, \beta, 0)$, $C(0, 0, \gamma)$ on the coordinate axes, its volume $V$ is given by $V = \frac{1}{6} |\alpha \beta \gamma|$. The absolute value ensures a positive volume.

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Step-by-Step Solution

Step 1: Determine the Normal Vector to the Plane P

The plane P is parallel to two lines with direction ratios $(-2, 1, -3)$ and $(-1, 2, -2)$. Let the direction vectors of these lines be $\vec{d_1} = -2\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{d_2} = -\hat{i} + 2\hat{j} - 2\hat{k}$.

• Why this step? A plane parallel to two lines has a normal vector that is perpendicular to the direction vectors of both lines. The cross product of two vectors yields a vector perpendicular to both.

We calculate the normal vector $\vec{n}$ using the cross product: $\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & -3 \\ -1 & 2 & -2 \end{vmatrix}$ $\vec{n} = \hat{i}((1)(-2) - (-3)(2)) - \hat{j}((-2)(-2) - (-3)(-1)) + \hat{k}((-2)(2) - (1)(-1))$ $\vec{n} = \hat{i}(-2 + 6) - \hat{j}(4 - 3) + \hat{k}(-4 + 1)$ $\vec{n} = 4\hat{i} - \hat{j} - 3\hat{k}$ Thus, the normal vector to the plane P is $\vec{n} = (4, -1, -3)$.

Step 2: Find the Equation of the Plane P

The plane P contains the point $(2, 2, -2)$ and has a normal vector $\vec{n} = (4, -1, -3)$.

• Why this step? With a point on the plane and its normal vector, we can directly write down the equation of the plane.

Using the formula $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$: $4(x-2) - 1(y-2) - 3(z-(-2)) = 0$ $4(x-2) - (y-2) - 3(z+2) = 0$ Expand and simplify the equation: $4x - 8 - y + 2 - 3z - 6 = 0$ $4x - y - 3z - 12 = 0$ $4x - y - 3z = 12$ This is the standard form of the equation of plane P.

Step 3: Determine the Intercepts $\alpha, \beta, \gamma$

To find the intercepts, we convert the plane's equation into the intercept form $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$.

• Why this step? The problem asks for the intercepts, which are directly identifiable from the intercept form.

Divide the entire equation $4x - y - 3z = 12$ by 12: $\frac{4x}{12} - \frac{y}{12} - \frac{3z}{12} = \frac{12}{12}$ $\frac{x}{3} + \frac{y}{-12} + \frac{z}{-4} = 1$ By comparing this with the intercept form, we identify the intercepts: $\alpha = 3$ $\beta = -12$ $\gamma = -4$

Step 4: Calculate the value of p

The problem defines $p = \alpha + \beta + \gamma$.

• Why this step? This is a direct calculation based on the definition given in the question.

Substitute the values of $\alpha, \beta, \gamma$: $p = 3 + (-12) + (-4)$ $p = 3 - 12 - 4$ $p = -13$

Step 5: Calculate the Volume V of the Tetrahedron OABC

The volume $V$ of the tetrahedron $OABC$ (where $O$ is the origin and $A, B, C$ are the points where the plane intersects the coordinate axes) is given by the formula $V = \frac{1}{6} |\alpha \beta \gamma|$.

• Why this step? This is the standard formula for the volume of a tetrahedron with vertices at the origin and on the coordinate axes. The absolute value is crucial as volume must be positive.

Substitute the values of $\alpha, \beta, \gamma$: $V = \frac{1}{6} |(3)(-12)(-4)|$ $V = \frac{1}{6} |(3)(48)|$ $V = \frac{1}{6} |144|$ $V = \frac{144}{6}$ $V = 24$

Step 6: Form the Ordered Pair (V, p)

We have calculated $V = 24$ and $p = -13$. The ordered pair $(V, p)$ is $(24, -13)$.

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Common Mistakes & Tips

• Absolute Value for Volume: Always remember to take the absolute value of the product $\alpha \beta \gamma$ when calculating the volume of the tetrahedron, as volume is a positive quantity.
• Sign Errors: Be careful with signs during the cross product calculation and when expanding the plane equation. A small sign error can lead to incorrect intercepts.
• Intercept Form Conversion: Ensure the plane equation is correctly converted to the intercept form $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 1$. This includes correctly handling negative signs in the denominator (e.g., $\frac{-y}{12}$ should be written as $\frac{y}{-12}$).

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Summary

First, we determined the normal vector to the plane by taking the cross product of the direction vectors of the two parallel lines. Using this normal vector and the given point on the plane, we found the equation of the plane. We then converted this equation into the intercept form to find the intercepts $\alpha, \beta, \gamma$. With these intercepts, we calculated $p = \alpha + \beta + \gamma$ and the volume $V = \frac{1}{6} |\alpha \beta \gamma|$ of the tetrahedron OABC. The calculated values are $V=24$ and $p=-13$.

The final answer is $\boxed{(24,-13)}$, which corresponds to option (B).