1. Key Concepts and Formulas

• Equation of a Plane through the Line of Intersection of Two Planes: If $P_1: A_1 x + B_1 y + C_1 z + D_1 = 0$ and $P_2: A_2 x + B_2 y + C_2 z + D_2 = 0$ are the equations of two planes, then the equation of any plane passing through their line of intersection is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar parameter. This represents a family of planes.
• Normal Vector of a Plane: For a plane with equation $Ax + By + Cz + D = 0$, the vector $\vec{N} = \langle A, B, C \rangle$ is its normal vector, which is perpendicular to the plane.
• Direction Vector of a Line: For a line given by $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$, the vector $\vec{d} = \langle l, m, n \rangle$ is its direction vector, which is parallel to the line.
• Condition for Parallelism between a Plane and a Line: A plane is parallel to a line if and only if the normal vector of the plane is perpendicular to the direction vector of the line. Mathematically, this means their dot product is zero: $\vec{N} \cdot \vec{d} = 0$.

2. Step-by-Step Solution

Step 1: Formulate the Equation of the Family of Planes We are given two planes whose line of intersection the required plane passes through. Let $P_1$ be $2x - y + z - 3 = 0$ and $P_2$ be $4x - 3y + 5z + 9 = 0$. According to the key concept, the equation of any plane passing through their line of intersection is $P_1 + \lambda P_2 = 0$. Substituting the equations of $P_1$ and $P_2$: $(2x - y + z - 3) + \lambda (4x - 3y + 5z + 9) = 0$ To easily identify the normal vector components, we rearrange this equation into the standard form $Ax + By + Cz + D = 0$: $(2 + 4\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda)z + (-3 + 9\lambda) = 0 \quad \text{...(i)}$ The normal vector to this plane, $\vec{N}$, is $\langle (2+4\lambda), (-1-3\lambda), (1+5\lambda) \rangle$.

Step 2: Identify the Direction Vector of the Given Line The problem states that the required plane is parallel to the line $\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$. From the standard form of a line's equation, the direction ratios of this line are $\langle -2, 4, 5 \rangle$. Therefore, the direction vector of the line, $\vec{d}$, is $\langle -2, 4, 5 \rangle$.

Step 3: Apply the Parallelism Condition to Find $\lambda$ Since the plane (i) is parallel to the given line, its normal vector $\vec{N}$ must be perpendicular to the line's direction vector $\vec{d}$. This means their dot product must be zero: $\vec{N} \cdot \vec{d} = 0$. Substituting the components of $\vec{N}$ and $\vec{d}$: $(2+4\lambda)(-2) + (-1-3\lambda)(4) + (1+5\lambda)(5) = 0$ Now, we expand and simplify the equation to solve for $\lambda$: $-4 - 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$ Combine the constant terms and the terms involving $\lambda$: $(-4 - 4 + 5) + (-8\lambda - 12\lambda + 25\lambda) = 0$ $-3 + 5\lambda = 0$ $5\lambda = 3$ $\lambda = \frac{3}{5}$

Step 4: Determine the Equation of the Specific Plane Now that we have the value of $\lambda$, we substitute it back into the equation of the family of planes (i) to find the specific equation of the required plane: $(2 + 4\left(\frac{3}{5}\right))x + (-1 - 3\left(\frac{3}{5}\right))y + (1 + 5\left(\frac{3}{5}\right))z + (-3 + 9\left(\frac{3}{5}\right)) = 0$ $\left(2 + \frac{12}{5}\right)x + \left(-1 - \frac{9}{5}\right)y + (1 + 3)z + \left(-3 + \frac{27}{5}\right) = 0$ $\left(\frac{10+12}{5}\right)x + \left(\frac{-5-9}{5}\right)y + 4z + \left(\frac{-15+27}{5}\right) = 0$ $\frac{22}{5}x - \frac{14}{5}y + 4z + \frac{12}{5} = 0$ To eliminate the fractions and simplify, multiply the entire equation by 5: $5 \left( \frac{22}{5}x - \frac{14}{5}y + 4z + \frac{12}{5} \right) = 5 \cdot 0$ $22x - 14y + 20z + 12 = 0$ The problem states that the equation of the plane is $ax+by+cz+6=0$. Our derived equation has a constant term of 12. To match the constant term to 6, we divide our entire equation by 2: $\frac{22x - 14y + 20z + 12}{2} = \frac{0}{2}$ $11x - 7y + 10z + 6 = 0$ This equation is now in the form $ax+by+cz+6=0$.

Step 5: Calculate $a+b+c$ By comparing our derived equation $11x - 7y + 10z + 6 = 0$ with the given form $ax+by+cz+6=0$, we can identify the coefficients: $a = 11$ $b = -7$ $c = 10$ Finally, we calculate the sum $a+b+c$: $a+b+c = 11 + (-7) + 10$ $a+b+c = 11 - 7 + 10$ $a+b+c = 4 + 10$ $a+b+c = 14$

3. Common Mistakes & Tips

• Sign Errors: Pay close attention to negative signs, especially when expanding expressions involving $\lambda$ and combining like terms.
• Distinguishing Normal and Direction Vectors: Remember that the plane's normal vector is perpendicular to the plane, while the line's direction vector is parallel to the line. The parallelism condition $\vec{N} \cdot \vec{d} = 0$ is for when the plane is parallel to the line.
• Matching Constant Terms: Always compare your derived plane equation with the target form given in the question. If the constant terms differ, scale your entire equation (by multiplying or dividing) to match it before identifying $a, b, c$. Failing to do so will lead to incorrect coefficients.

4. Summary

This problem required us to find the equation of a plane that passes through the intersection of two given planes and is parallel to a given line. We began by setting up the general equation for a plane passing through the line of intersection using the family of planes concept, introducing a parameter $\lambda$. Then, we utilized the condition for parallelism between a plane and a line, which states that the normal vector of the plane must be perpendicular to the direction vector of the line. This allowed us to form an equation involving $\lambda$, which we solved. Substituting the value of $\lambda$ back into the general equation yielded the specific plane. Finally, we adjusted the constant term of our plane equation to match the specified form and then identified the coefficients $a, b, c$ to compute their sum, which was 14.

5. Final Answer

The final answer is $\boxed{14}$, which corresponds to option (C).