1. Key Concepts and Formulas

• 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 $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.
• Normal Vector Properties:
  • If a plane passes through two points, say $P_1$ and $P_2$, then the vector $\vec{P_1P_2}$ lies in the plane. The normal vector to the plane must be perpendicular to any vector lying in the plane.
  • If two planes are perpendicular, their respective normal vectors are also perpendicular.
• Cross Product: The cross product of two non-parallel vectors $\vec{u}$ and $\vec{v}$ yields a vector $\vec{u} \times \vec{v}$ that is perpendicular to both $\vec{u}$ and $\vec{v}$. This property is essential for finding a vector perpendicular to two given vectors.

2. Step-by-Step Solution

Step 1: Identify the given information and objective.

We need to find the value of $2\alpha - 3\beta$. This value is derived from the fact that the point $P(2, \alpha, \beta)$ lies on a specific plane (let's call it Plane P1). The conditions defining Plane P1 are:

• It passes through points $A(3, 4, 2)$ and $B(7, 0, 6)$.
• It is perpendicular to Plane P2, whose equation is $2x - 5y = 15$.

Step 2: Determine vectors relevant to finding the normal of Plane P1.

• Vector lying in Plane P1: Since Plane P1 passes through points $A(3, 4, 2)$ and $B(7, 0, 6)$, the vector $\vec{AB}$ lies within Plane P1. We calculate $\vec{AB}$: $\vec{AB} = B - A = (7 - 3, 0 - 4, 6 - 2) = (4, -4, 4)$ Explanation: Any vector connecting two points on a plane lies in that plane. The normal vector of Plane P1 must be perpendicular to $\vec{AB}$.

• Normal vector of Plane P2: The equation of Plane P2 is $2x - 5y + 0z = 15$. From the coefficients of $x, y, z$, we can directly identify its normal vector, $\vec{n_2}$. $\vec{n_2} = (2, -5, 0)$ Explanation: For a plane $Ax + By + Cz = D$, the vector $(A, B, C)$ is a normal vector to the plane.

Step 3: Calculate the normal vector of Plane P1 ($\vec{n_1}$).

Let $\vec{n_1}$ be the normal vector of Plane P1.

• Since Plane P1 is perpendicular to Plane P2, their normal vectors must be perpendicular. Thus, $\vec{n_1} \perp \vec{n_2}$.
• Since $\vec{AB}$ lies in Plane P1, $\vec{n_1}$ must be perpendicular to $\vec{AB}$.

Because $\vec{n_1}$ is perpendicular to both $\vec{n_2}$ and $\vec{AB}$, it must be parallel to their cross product. We can take $\vec{n_1}$ to be $\vec{AB} \times \vec{n_2}$ (or $\vec{n_2} \times \vec{AB}$, which would give a normal vector in the opposite direction but still valid).

Let's compute the cross product $\vec{AB} \times \vec{n_2}$: $\vec{n_1} = \vec{AB} \times \vec{n_2} = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 4 & -4 & 4 \\ 2 & -5 & 0 \end{vmatrix}$ $= \widehat i((-4)(0) - (4)(-5)) - \widehat j((4)(0) - (4)(2)) + \widehat k((4)(-5) - (-4)(2))$ $= \widehat i(0 - (-20)) - \widehat j(0 - 8) + \widehat k(-20 - (-8))$ $= 20\widehat i + 8\widehat j - 12\widehat k$ So, a normal vector for Plane P1 is $\vec{n_1} = (20, 8, -12)$.

Explanation: The cross product of two vectors produces a vector perpendicular to both. This resulting vector serves as the normal vector for Plane P1. Tip: We can simplify this normal vector by dividing by a common scalar. Dividing by $4$ gives a simpler normal vector $(5, 2, -3)$, which will make subsequent calculations easier. Both $(20, 8, -12)$ and $(5, 2, -3)$ are valid normal vectors for the same plane. Let's use $\vec{n_1} = (5, 2, -3)$.

Step 4: Formulate the Equation of Plane P1.

We use the general equation of a plane $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$. Using the simplified normal vector $\vec{n_1} = (5, 2, -3)$ and point $A(3, 4, 2)$ which lies on Plane P1: $5(x - 3) + 2(y - 4) - 3(z - 2) = 0$ $5x - 15 + 2y - 8 - 3z + 6 = 0$ $5x + 2y - 3z - 17 = 0$ Explanation: We substitute the components of the normal vector $(A, B, C)$ and the coordinates of a point $(x_0, y_0, z_0)$ known to be on the plane into the plane equation formula.

Step 5: Use the given point $(2, \alpha, \beta)$ to find the relationship between $\alpha$ and $\beta$.

The point $(2, \alpha, \beta)$ lies on Plane P1. Therefore, its coordinates must satisfy the equation of Plane P1: $5(2) + 2(\alpha) - 3(\beta) - 17 = 0$ $10 + 2\alpha - 3\beta - 17 = 0$ $2\alpha - 3\beta - 7 = 0$ $2\alpha - 3\beta = 7$ Explanation: If a point lies on a plane, plugging its coordinates into the plane's equation must result in a true statement (i.e., the equation holds).

Step 6: Calculate the required expression.

The problem asks for the value of $2\alpha - 3\beta$. From Step 5, we found: $2\alpha - 3\beta = 7$ Explanation: The final step is a direct retrieval of the value determined in the previous step.

3. Common Mistakes & Tips

• Sign Errors: Be very careful with signs when calculating cross products and expanding the plane equation. A single sign error can lead to an incorrect normal vector or plane equation.
• Vector Direction: While $\vec{AB} \times \vec{n_2}$ and $\vec{n_2} \times \vec{AB}$ result in normal vectors pointing in opposite directions, both are valid normal vectors for the same plane. Choosing one consistently is important.
• Simplifying Normal Vector: Always simplify the normal vector by dividing by the greatest common divisor of its components. This reduces the magnitude of numbers and makes calculations less error-prone.
• Constant Term D: Remember to correctly calculate the constant term $D$ in the plane equation $Ax + By + Cz + D = 0$ by substituting a known point on the plane.

4. Summary

To determine the value of $2\alpha - 3\beta$, we first established the equation of Plane P1. This involved finding two vectors perpendicular to Plane P1's normal vector: one from the two given points on the plane ($\vec{AB}$) and another from the normal of the perpendicular plane ($\vec{n_2}$). Their cross product provided the normal vector for Plane P1. Using this normal vector and one of the given points, we constructed the equation of Plane P1. Finally, we substituted the coordinates of the point $(2, \alpha, \beta)$ into the plane's equation to derive the required relationship and value.

5. Final Answer

The value of $2\alpha - 3\beta$ is 7.

The final answer is $\boxed{7}$ which corresponds to option (B).