1. Key Concepts and Formulas

This problem requires us to determine the equation of a plane based on specific geometric conditions. The fundamental concepts and formulas we'll utilize are:

• Equation of a Plane Passing Through a Given Point: The general equation of a plane that passes through a point $(x_1, y_1, z_1)$ and has a normal vector with direction ratios $(A, B, C)$ is given by: $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$ Here, $(A, B, C)$ represents the direction ratios of any vector perpendicular to the plane (its normal vector).
• Normal Vector of a Plane: For a plane defined by the equation $Ax + By + Cz + D = 0$, its normal vector is $\vec{n} = A\mathbf{i} + B\mathbf{j} + C\mathbf{k}$, or simply $(A, B, C)$.
• Condition for Perpendicular Planes: Two planes are perpendicular if and only if their respective normal vectors are perpendicular. Mathematically, if $\vec{n_1}$ and $\vec{n_2}$ are the normal vectors of two planes, then for them to be perpendicular, their dot product must be zero: $\vec{n_1} \cdot \vec{n_2} = 0$
• Finding a Vector Perpendicular to Two Given Vectors (Cross Product): If a vector $\vec{v}$ is perpendicular to two non-parallel vectors $\vec{u_1}$ and $\vec{u_2}$, then $\vec{v}$ must be parallel to their cross product. That is, $\vec{v} = k (\vec{u_1} \times \vec{u_2})$ for some non-zero scalar $k$. The cross product of two vectors $\vec{u_1} = (u_x, u_y, u_z)$ and $\vec{u_2} = (v_x, v_y, v_z)$ is given by: $\vec{u_1} \times \vec{u_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix} = (u_y v_z - u_z v_y)\mathbf{i} - (u_x v_z - u_z v_x)\mathbf{j} + (u_x v_y - u_y v_x)\mathbf{k}$

2. Step-by-Step Solution

Step 1: Set up the General Equation of the Required Plane

We are given that the required plane passes through the point $(-2, 3, 5)$. Let the direction ratios of its normal vector be $(a, b, c)$. Using the formula for the equation of a plane passing through a given point, the equation of our plane can be written as: $a(x - (-2)) + b(y - 3) + c(z - 5) = 0$ $\Rightarrow a(x + 2) + b(y - 3) + c(z - 5) = 0 \quad \text{...(1)}$ Our primary objective is to determine the values of $a, b,$ and $c$.

Step 2: Identify Normal Vectors of the Given Perpendicular Planes

The problem states that the required plane is perpendicular to two other planes:

1. $2x + 4y + 5z = 8$
2. $3x - 2y + 3z = 5$

The normal vector of a plane $Ax + By + Cz + D = 0$ is $(A, B, C)$. So, the normal vector of the first given plane is $\vec{n_1} = (2, 4, 5)$. And the normal vector of the second given plane is $\vec{n_2} = (3, -2, 3)$.

Step 3: Determine the Normal Vector of the Required Plane using the Cross Product

Since the required plane is perpendicular to both the first and second given planes, its normal vector, $\vec{n} = (a, b, c)$, must be perpendicular to both $\vec{n_1}$ and $\vec{n_2}$. A vector perpendicular to two given vectors is parallel to their cross product. Therefore, we can find the direction ratios $(a, b, c)$ by calculating the cross product of $\vec{n_1}$ and $\vec{n_2}$: $\vec{n} = \vec{n_1} \times \vec{n_2}$ $\vec{n} = (2\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) \times (3\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})$ Calculating the cross product: $\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 4 & 5 \\ 3 & -2 & 3 \end{vmatrix}$ $\vec{n} = \mathbf{i}((4)(3) - (5)(-2)) - \mathbf{j}((2)(3) - (5)(3)) + \mathbf{k}((2)(-2) - (4)(3))$ $\vec{n} = \mathbf{i}(12 - (-10)) - \mathbf{j}(6 - 15) + \mathbf{k}(-4 - 12)$ $\vec{n} = \mathbf{i}(22) - \mathbf{j}(-9) + \mathbf{k}(-16)$ $\vec{n} = 22\mathbf{i} + 9\mathbf{j} - 16\mathbf{k}$ Thus, the direction ratios of the normal vector to the required plane are $(a, b, c) = (22, 9, -16)$.

Step 4: Substitute the Normal Vector into the Plane Equation

Now, substitute the values of $a = 22, b = 9, c = -16$ back into Equation (1): $22(x + 2) + 9(y - 3) + (-16)(z - 5) = 0$ $22(x + 2) + 9(y - 3) - 16(z - 5) = 0$ Expand and simplify the equation: $22x + 44 + 9y - 27 - 16z + 80 = 0$ Combine the constant terms: $22x + 9y - 16z + (44 - 27 + 80) = 0$ $22x + 9y - 16z + (17 + 80) = 0$ $22x + 9y - 16z + 97 = 0$

Step 5: Compare with the Given Form and Identify $\alpha, \beta, \gamma$

The problem states that the equation of the plane is given in the form $\alpha x+\beta y+\gamma z+97=0$. Comparing our derived equation $22x + 9y - 16z + 97 = 0$ with the given form, we can directly identify the coefficients: $\alpha = 22$ $\beta = 9$ $\gamma = -16$

Step 6: Calculate the Final Value $\alpha+\beta+\gamma$

Finally, we need to compute the sum $\alpha+\beta+\gamma$: $\alpha + \beta + \gamma = 22 + 9 + (-16)$ $\alpha + \beta + \gamma = 31 - 16$ $\alpha + \beta + \gamma = 15$

3. Common Mistakes & Tips

• Sign Errors in Cross Product/Determinants: Be extremely meticulous with signs when calculating the cross product or determinants. A single sign error can lead to incorrect direction ratios and thus an incorrect plane equation.
• Understanding Direction Ratios: Remember that $(a, b, c)$ are direction ratios, meaning any non-zero scalar multiple $(ka, kb, kc)$ represents the same normal direction. However, in this problem, the constant term +97 in the target equation αx + βy + γz + 97 = 0 fixes the overall scalar multiple of the equation, ensuring unique values for $\alpha, \beta, \gamma$.
• Verification: After finding the equation of the plane, it's good practice to quickly verify if it satisfies the given conditions:
  1. Point on the plane: Substitute $(-2,3,5)$ into $22x + 9y - 16z + 97 = 0$: $22(-2) + 9(3) - 16(5) + 97 = -44 + 27 - 80 + 97 = -17 - 80 + 97 = -97 + 97 = 0$. This is correct.
  2. Perpendicularity to other planes: Check if the normal vector $(22, 9, -16)$ is perpendicular to $(2, 4, 5)$ and $(3, -2, 3)$ by taking dot products:
     • $(22, 9, -16) \cdot (2, 4, 5) = (22)(2) + (9)(4) + (-16)(5) = 44 + 36 - 80 = 80 - 80 = 0$. Correct.
     • $(22, 9, -16) \cdot (3, -2, 3) = (22)(3) + (9)(-2) + (-16)(3) = 66 - 18 - 48 = 48 - 48 = 0$. Correct.

4. Summary

This problem is a standard application of 3D geometry principles to find the equation of a plane. The key steps involved identifying the normal vectors of the given planes, using their cross product to find the normal vector of the required plane, and then using the given point to form the complete plane equation. Finally, by comparing the derived equation with the specified format, the coefficients $\alpha, \beta, \gamma$ were determined, leading to the required sum.

5. Final Answer

The calculated value of $\alpha+\beta+\gamma$ is 15.

The final answer is $\boxed{15}$ which corresponds to option (A).