1. Key Concepts and Formulas

• Condition for a Line to Lie in a Plane: For a line to lie entirely within a plane, two conditions must be simultaneously satisfied:
  1. Point Condition: Any arbitrary point on the line must also lie in the plane. This means that if $(x_1, y_1, z_1)$ is a point on the line, it must satisfy the plane's equation $Ax_1+By_1+Cz_1+D=0$.
  2. Perpendicularity Condition: The direction vector of the line must be perpendicular (orthogonal) to the normal vector of the plane. This implies their dot product must be zero: $\vec{b} \cdot \vec{n} = 0$.
• Extracting Information from Equations:
  • From the symmetric form of a line $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$:
    • A point on the line is $(x_1, y_1, z_1)$.
    • The direction vector of the line is $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.
  • From the general equation of a plane $Ax+By+Cz+D=0$:
    • The normal vector to the plane is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

2. Step-by-Step Solution

Step 1: Extract the point and direction vector of the line, and the normal vector of the plane.

• What we are doing: We are identifying the key geometric components (a point and direction vector for the line, and a normal vector for the plane) from their given equations.
• Why we are doing it: These components are essential for applying the conditions for a line to lie in a plane.
• Action:
  • Given line: $\frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 2}{2}$
    • By comparing with $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, we find:
    • A point on the line is $P_0(x_1, y_1, z_1) = (2, 1, -2)$. (Note: $z+2 = z-(-2)$).
    • The direction vector of the line is $\vec{b} = 3\hat{i} - 5\hat{j} + 2\hat{k}$.
  • Given plane: $x + 3y - \alpha z + \beta = 0$
    • By comparing with $Ax+By+Cz+D=0$, we find:
    • The normal vector to the plane is $\vec{n} = 1\hat{i} + 3\hat{j} - \alpha\hat{k}$.
• Reasoning: Correctly identifying these vectors and points is the foundational step before applying any conditions.

Step 2: Apply the Perpendicularity Condition to find $\alpha$.

• What we are doing: We are using the condition that the direction vector of the line must be perpendicular to the normal vector of the plane.
• Why we are doing it: If the line lies in the plane, its direction must be parallel to the plane, and thus perpendicular to the plane's normal vector. This condition typically allows us to solve for one of the unknowns directly.
• Action: For $\vec{b}$ and $\vec{n}$ to be perpendicular, their dot product must be zero ($\vec{b} \cdot \vec{n} = 0$).
• Calculation: $(3)(1) + (-5)(3) + (2)(-\alpha) = 0$ $3 - 15 - 2\alpha = 0$ $-12 - 2\alpha = 0$ $-2\alpha = 12$ $\alpha = -6$
• Reasoning: This calculation directly provides the value of $\alpha$, as it's the only unknown in this condition.

Step 3: Apply the Point Condition to find $\beta$.

• What we are doing: We are using the condition that the point $P_0(2, 1, -2)$ (which lies on the line) must also lie in the plane.
• Why we are doing it: Since the line lies entirely within the plane, every point on the line, including $P_0$, must satisfy the plane's equation. This condition will allow us to find the remaining unknown, $\beta$.
• Action: Substitute the coordinates of $P_0(2, 1, -2)$ and the value of $\alpha = -6$ into the plane equation $x + 3y - \alpha z + \beta = 0$.
• Calculation: $(2) + 3(1) - (-6)(-2) + \beta = 0$ $2 + 3 - (12) + \beta = 0$ $5 - 12 + \beta = 0$ $-7 + \beta = 0$ $\beta = 7$
• Reasoning: With $\alpha$ determined, substituting it along with the point coordinates into the plane equation allows us to isolate and solve for $\beta$.

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful when extracting the coordinates of the point from the line equation (e.g., $z+2$ implies $z_1 = -2$) and the coefficients of the normal vector from the plane equation (e.g., $-\alpha z$ implies the z-component is $-\alpha$).
• Understanding Conditions: Ensure you clearly understand why these two conditions are necessary and sufficient. A line could be parallel to a plane (satisfying the dot product condition) but not lie in it (failing the point condition).
• Vector Notation: Be comfortable with both component form and $\hat{i}, \hat{j}, \hat{k}$ notation for vectors and their dot product.

4. Summary

To determine the unknown parameters when a line lies within a plane, we systematically apply two fundamental geometric conditions. First, we utilize the fact that the line's direction vector must be perpendicular to the plane's normal vector, which allows us to solve for $\alpha$. Second, we ensure that any point on the line satisfies the plane's equation, enabling us to solve for $\beta$ using the value of $\alpha$ already found. Following these steps, we determined $\alpha = -6$ and $\beta = 7$.

5. Final Answer

The values are $\alpha = -6$ and $\beta = 7$. Thus, $(\alpha, \beta) = (-6, 7)$.

The final answer is $\boxed{(-6, 7)}$ which corresponds to option (A).