Here's a clear, educational, and well-structured solution to the problem.

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1. Key Concepts and Formulas

• Family of Planes: The equation of any plane passing through the line of intersection of two planes $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$ is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is an arbitrary real constant. This means $(A_1x + B_1y + C_1z + D_1) + \lambda (A_2x + B_2y + C_2z + D_2) = 0$.
• Condition for a Specific Plane: To find a unique plane from this family, an additional condition is required, such as the plane passing through a given point. Substituting the coordinates of the point into the family of planes equation allows us to determine the specific value of $\lambda$.
• General Equation of a Plane: The general equation of a plane is $Ax + By + Cz + D = 0$, where $A, B, C$ are the direction ratios of the normal to the plane, and $D$ is a constant.

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2. Step-by-Step Solution

Step 1: Formulate the Equation of the Family of Planes We are given the equations of two planes: Plane 1 ($P_1$): $2x - 7y + 4z - 3 = 0$ Plane 2 ($P_2$): $3x - 5y + 4z + 11 = 0$

The equation of any plane passing through their line of intersection is given by $P_1 + \lambda P_2 = 0$. Substituting the given plane equations: $(2x - 7y + 4z - 3) + \lambda (3x - 5y + 4z + 11) = 0 \quad (*)$ Explanation: This equation represents an infinite set of planes, all of which contain the line where $P_1$ and $P_2$ intersect. The parameter $\lambda$ distinguishes these different planes.

Step 2: Use the Given Point to Determine $\lambda$ We are told that the desired plane passes through the point $(-2, 1, 3)$. Since the plane must pass through this point, its coordinates must satisfy the plane's equation. Substitute $x = -2$, $y = 1$, $z = 3$ into equation $(*)$: $(2(-2) - 7(1) + 4(3) - 3) + \lambda (3(-2) - 5(1) + 4(3) + 11) = 0$ Explanation: This step allows us to find the unique value of $\lambda$ that corresponds to the specific plane that passes through the given point.

Step 3: Solve for $\lambda$ Now, simplify the expression and solve for $\lambda$: $(-4 - 7 + 12 - 3) + \lambda (-6 - 5 + 12 + 11) = 0$ Combine the terms within each parenthesis: $(-11 + 12 - 3) + \lambda (-11 + 12 + 11) = 0$ $(-2) + \lambda (12) = 0$ $12\lambda = 2$ $\lambda = \frac{2}{12}$ $\lambda = \frac{1}{6}$ Explanation: The value $\lambda = \frac{1}{6}$ uniquely identifies the plane we are looking for from the family of planes.

Step 4: Find the Equation of the Specific Plane Substitute the value of $\lambda = \frac{1}{6}$ back into equation $(*)$: $(2x - 7y + 4z - 3) + \frac{1}{6} (3x - 5y + 4z + 11) = 0$ To eliminate the fraction and simplify the equation, multiply the entire equation by 6: $6(2x - 7y + 4z - 3) + 1(3x - 5y + 4z + 11) = 0$ $12x - 42y + 24z - 18 + 3x - 5y + 4z + 11 = 0$ Now, group and combine the like terms (coefficients of $x$, $y$, $z$, and constant terms): $(12x + 3x) + (-42y - 5y) + (24z + 4z) + (-18 + 11) = 0$ $15x - 47y + 28z - 7 = 0$ Explanation: This is the unique equation of the plane that satisfies both conditions: passing through the line of intersection and passing through the point $(-2, 1, 3)$.

Step 5: Identify the Coefficients $a, b, c$ The problem states that the equation of the plane is $ax + by + cz - 7 = 0$. Comparing our derived equation $15x - 47y + 28z - 7 = 0$ with the given form, we can directly identify the coefficients: $a = 15$ $b = -47$ $c = 28$ Explanation: The constant term in our derived equation is already $-7$, which perfectly matches the form given in the problem statement. This means no further scaling of the equation is required to find $a, b, c$.

Step 6: Calculate the Value of the Expression $2a + b + c - 7$ Now, substitute the identified values of $a, b, c$ into the expression: $2a + b + c - 7 = 2(15) + (-47) + (28) - 7$ $= 30 - 47 + 28 - 7$ Group positive and negative terms for easier calculation: $= (30 + 28) - (47 + 7)$ $= 58 - 54$ $= 1$ Explanation: This is the final value requested by the problem, obtained by substituting the determined coefficients into the given algebraic expression.

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3. Common Mistakes & Tips

• Algebraic Errors: Be extremely cautious with signs and calculations, especially when substituting coordinates and combining terms. A single arithmetic mistake can propagate and lead to an incorrect value of $\lambda$ and consequently incorrect coefficients $a, b, c$.
• Constant Term Matching: Always ensure that the constant term in your final plane equation matches the form specified in the problem (e.g., $ax + by + cz - 7 = 0$). If your derived equation is, for example, $30x - 94y + 56z - 14 = 0$, you would divide the entire equation by 2 to get $15x - 47y + 28z - 7 = 0$ before identifying $a, b, c$.
• Understanding $\lambda$: Remember that $\lambda$ is a scalar parameter. Its value (positive, negative, or fractional) determines the specific plane within the family that satisfies the additional given condition.

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4. Summary

This problem effectively demonstrates the application of the "family of planes" concept in 3D Geometry. By combining the equations of two intersecting planes with a parameter $\lambda$, we form a general equation for any plane passing through their line of intersection. The specific value of $\lambda$ is then determined by using the additional condition that the plane passes through a given point. Once $\lambda$ is found, substituting it back yields the unique equation of the desired plane. Finally, by comparing this equation with the specified form, we identify the coefficients and evaluate the required expression.

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The final answer is $\boxed{1}$.