1. Key Concepts and Formulas

• Family of Planes: The equation of any plane that passes through the line of intersection of two given planes, $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$, can be represented as a linear combination of their equations: $P_1 + \lambda P_2 = 0$ or $(A_1x + B_1y + C_1z + D_1) + \lambda (A_2x + B_2y + C_2z + D_2) = 0$ Here, $\lambda$ is a scalar constant. Each value of $\lambda$ corresponds to a unique plane in this family.
• Point on a Plane: If a point $(x_0, y_0, z_0)$ lies on a plane $Ax + By + Cz + D = 0$, then its coordinates must satisfy the plane's equation, i.e., $Ax_0 + By_0 + Cz_0 + D = 0$. This condition is used to determine the specific value of $\lambda$.
• Unique Plane Representation: The equation of a plane is unique up to a non-zero scalar multiple. For example, $x+y+z+1=0$ and $2x+2y+2z+2=0$ represent the same plane. This allows us to scale an equation to match a given form.

2. Step-by-Step Solution

Step 1: Formulate the equation of the family of planes. We are given two planes:

• Plane 1 ($P_1$): $3x - 2y + 4z - 7 = 0$
• Plane 2 ($P_2$): $x + 5y - 2z + 9 = 0$

The equation of any plane containing their line of intersection is given by $P_1 + \lambda P_2 = 0$. Substituting the equations of $P_1$ and $P_2$: $(3x - 2y + 4z - 7) + \lambda (x + 5y - 2z + 9) = 0$ Reasoning: This equation represents all possible planes that pass through the common line where $P_1$ and $P_2$ intersect. Our goal is to find the specific value of $\lambda$ that defines the plane we are looking for.

Step 2: Simplify the general equation of the family. To make it easier to work with, we expand the equation and group terms by $x$, $y$, $z$, and the constant: $3x - 2y + 4z - 7 + \lambda x + 5\lambda y - 2\lambda z + 9\lambda = 0$ $(3 + \lambda)x + (-2 + 5\lambda)y + (4 - 2\lambda)z + (-7 + 9\lambda) = 0$ Reasoning: This rearrangement into the standard form $Ax+By+Cz+D=0$ prepares the equation for easy substitution of point coordinates in the next step.

Step 3: Use the given point to determine $\lambda$. The problem states that the desired plane passes through the point $(1, 4, -3)$. For this point to lie on the plane, its coordinates must satisfy the plane's equation. We substitute $x=1$, $y=4$, and $z=-3$ into the simplified equation from Step 2: $(3 + \lambda)(1) + (-2 + 5\lambda)(4) + (4 - 2\lambda)(-3) + (-7 + 9\lambda) = 0$ Reasoning: This substitution converts the equation of the plane, which contains variables $x, y, z$ and the unknown parameter $\lambda$, into an algebraic equation solely in terms of $\lambda$. Solving this equation will yield the unique value of $\lambda$ that identifies our specific plane.

Step 4: Solve for $\lambda$. Now, we simplify and solve the equation for $\lambda$: $(3 + \lambda) + (-8 + 20\lambda) + (-12 + 6\lambda) + (-7 + 9\lambda) = 0$ Combine the constant terms: $3 - 8 - 12 - 7 = -24$ Combine the terms containing $\lambda$: $\lambda + 20\lambda + 6\lambda + 9\lambda = 36\lambda$ The equation becomes: $36\lambda - 24 = 0$ Solving for $\lambda$: $36\lambda = 24$ $\lambda = \frac{24}{36}$ $\lambda = \frac{2}{3}$ Reasoning: Careful algebraic manipulation is crucial here to find the correct value of $\lambda$. This value uniquely defines the specific plane we are seeking.

Step 5: Determine the specific plane's equation. Substitute $\lambda = \frac{2}{3}$ back into the simplified plane equation from Step 2: $(3 + \lambda)x + (-2 + 5\lambda)y + (4 - 2\lambda)z + (-7 + 9\lambda) = 0$ Let's calculate each coefficient:

• Coefficient of $x$: $3 + \frac{2}{3} = \frac{9+2}{3} = \frac{11}{3}$
• Coefficient of $y$: $-2 + 5\left(\frac{2}{3}\right) = -2 + \frac{10}{3} = \frac{-6+10}{3} = \frac{4}{3}$
• Coefficient of $z$: $4 - 2\left(\frac{2}{3}\right) = 4 - \frac{4}{3} = \frac{12-4}{3} = \frac{8}{3}$
• Constant term: $-7 + 9\left(\frac{2}{3}\right) = -7 + 3(2) = -7 + 6 = -1$

So, the equation of the plane is: $\frac{11}{3}x + \frac{4}{3}y + \frac{8}{3}z - 1 = 0$ To eliminate fractions, we multiply the entire equation by 3: $11x + 4y + 8z - 3 = 0$ Reasoning: Substituting the determined value of $\lambda$ into the general equation yields the unique equation of the plane that satisfies both conditions: containing the line of intersection and passing through the given point. Multiplying by 3 simplifies the coefficients without changing the plane itself.

Step 6: Compare with the target equation and find $\alpha, \beta, \gamma$. The problem states that the equation of the plane is $\alpha x + \beta y + \gamma z + 3 = 0$. Our derived equation is $11x + 4y + 8z - 3 = 0$.

To match the constant term of $+3$ in the target equation, we need to multiply our derived equation by a factor of $-1$: $-1(11x + 4y + 8z - 3) = 0$ $-11x - 4y - 8z + 3 = 0$ Now, comparing this equation with $\alpha x + \beta y + \gamma z + 3 = 0$:

• $\alpha = -11$
• $\beta = -4$
• $\gamma = -8$ Reasoning: Since a plane's equation is unique up to a scalar multiple, we adjust our equation to match the constant term of the given form. This allows for direct identification of the coefficients $\alpha$, $\beta$, and $\gamma$.

Step 7: Calculate $\alpha + \beta + \gamma$. Finally, we compute the required sum: $\alpha + \beta + \gamma = (-11) + (-4) + (-8)$ $\alpha + \beta + \gamma = -11 - 4 - 8$ $\alpha + \beta + \gamma = -23$

3. Common Mistakes & Tips

• Arithmetic Errors: Be extremely vigilant with calculations, especially sign conventions, when expanding terms and solving for $\lambda$. A small error here propagates through the rest of the solution.
• Forgetting to Scale: Remember that the equation of a plane is unique only up to a non-zero scalar multiple. Always compare your final derived plane equation with the target form (e.g., $\alpha x + \beta y + \gamma z + D_{target} = 0$) and scale your equation accordingly to match the constant term $D_{target}$.
• Conceptual Understanding: Ensure you understand that $P_1 + \lambda P_2 = 0$ represents an entire family of planes passing through the line of intersection. The additional point condition helps pinpoint the unique plane within this family.

4. Summary

This problem required us to find the equation of a plane that belongs to a "family of planes" passing through the line of intersection of two given planes, and also satisfies an additional condition of passing through a specific point. We first formulated the general equation of the family as $P_1 + \lambda P_2 = 0$. Then, we substituted the coordinates of the given point into this equation to solve for the unique value of $\lambda$. With $\lambda$ determined, we substituted it back into the general equation to obtain the specific plane's equation. Finally, we scaled this equation to match the constant term of the given target form $\alpha x + \beta y + \gamma z + 3 = 0$ to identify $\alpha, \beta, \gamma$ and calculate their sum.

The final answer is $\boxed{\text{-23}}$, which corresponds to option (A).