Key Concepts and Formulas

1. Normal Vector of a Plane: For a plane given by the equation $Ax + By + Cz + D = 0$, the vector $\vec{n} = A\widehat{i} + B\widehat{j} + C\widehat{k}$ is its normal vector. This vector is perpendicular to the plane and defines its orientation in 3D space.

2. Perpendicular Planes and Cross Product: If a plane $P$ is perpendicular to two other planes $P_1$ and $P_2$, then the normal vector of plane $P$, say $\vec{n}$, must be perpendicular to the normal vectors of $P_1$ ($\vec{n_1}$) and $P_2$ ($\vec{n_2}$). The cross product of $\vec{n_1}$ and $\vec{n_2}$ (i.e., $\vec{n_1} \times \vec{n_2}$) yields a vector that is mutually orthogonal to both $\vec{n_1}$ and $\vec{n_2}$. Thus, $\vec{n}$ can be taken as $\vec{n_1} \times \vec{n_2}$ (or any non-zero scalar multiple thereof).

3. Equation of a Plane (Point-Normal Form): The equation of a plane that passes through a point $(x_1, y_1, z_1)$ and has a normal vector $\vec{n} = A\widehat{i} + B\widehat{j} + C\widehat{k}$ is given by: $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$

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Step-by-Step Solution

Step 1: Identify the Normal Vectors of the Given Planes

We are given two planes to which our required plane is perpendicular:

• Plane 1: $2x + y - z = 2$
• Plane 2: $x - y - z = 3$

The normal vector of a plane $Ax + By + Cz + D = 0$ is obtained by taking the coefficients of $x, y, z$.

• For Plane 1, the normal vector is $\vec{n_1} = 2\widehat{i} + \widehat{j} - \widehat{k}$.
• For Plane 2, the normal vector is $\vec{n_2} = \widehat{i} - \widehat{j} - \widehat{k}$.

Why this step? The normal vectors are crucial because they encapsulate the orientation of the given planes. Since our target plane is perpendicular to these, its normal vector must be derived from $\vec{n_1}$ and $\vec{n_2}$.

Step 2: Determine the Normal Vector of the Required Plane

The problem states that the required plane is perpendicular to both Plane 1 and Plane 2. This implies that its normal vector, let's denote it as $\vec{n}$, must be perpendicular to both $\vec{n_1}$ and $\vec{n_2}$. The mathematical operation that yields a vector perpendicular to two given vectors is the cross product. So, we compute $\vec{n} = \vec{n_1} \times \vec{n_2}$: $\vec{n} = (2\widehat{i} + \widehat{j} - \widehat{k}) \times (\widehat{i} - \widehat{j} - \widehat{k})$ Using the determinant form for the cross product: $\vec{n} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 2 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix}$ Expanding the determinant: $\vec{n} = \widehat{i}((1)(-1) - (-1)(-1)) - \widehat{j}((2)(-1) - (-1)(1)) + \widehat{k}((2)(-1) - (1)(1))$ $\vec{n} = \widehat{i}(-1 - 1) - \widehat{j}(-2 + 1) + \widehat{k}(-2 - 1)$ $\vec{n} = \widehat{i}(-2) - \widehat{j}(-1) + \widehat{k}(-3)$ $\vec{n} = -2\widehat{i} + \widehat{j} - 3\widehat{k}$ Thus, the normal vector of the required plane is $\vec{n} = -2\widehat{i} + \widehat{j} - 3\widehat{k}$. This means the coefficients $(A, B, C)$ for our plane's equation are $(-2, 1, -3)$.

Why this step? The geometric condition of perpendicularity to two planes directly translates to its normal vector being perpendicular to the normal vectors of those two planes. The cross product is the precise mathematical tool to find such a vector.

Step 3: Formulate the Equation of the Required Plane

We now have all the necessary information to write the equation of the plane:

• It passes through the point $(x_1, y_1, z_1) = (-1, 0, -2)$.
• Its normal vector is $\vec{n} = -2\widehat{i} + \widehat{j} - 3\widehat{k}$, so $(A, B, C) = (-2, 1, -3)$.

Using the point-normal form of the plane equation, $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$: $-2(x - (-1)) + 1(y - 0) - 3(z - (-2)) = 0$ $-2(x + 1) + y - 3(z + 2) = 0$

Why this step? The point-normal form is the most direct way to construct the equation of a plane when a point on the plane and its normal vector are known.

Step 4: Simplify the Equation and Match the Given Form

Now, we expand and simplify the equation: $-2x - 2 + y - 3z - 6 = 0$ $-2x + y - 3z - 8 = 0$ The problem asks for the plane in the form $ax + by + cz + 8 = 0$. Our derived equation is $-2x + y - 3z - 8 = 0$. To match the constant term ($+8$), we multiply the entire equation by $-1$. Multiplying a plane equation by a non-zero scalar does not change the plane itself, only the representation of its coefficients. $(-1) \times (-2x + y - 3z - 8) = (-1) \times 0$ $2x - y + 3z + 8 = 0$

Why this step? Simplifying the equation brings it to a standard general form. Matching the constant term to the format specified in the question ($+8$) is essential for correctly identifying the coefficients $a, b, c$.

Step 5: Identify the Coefficients $a, b, c$ and Calculate $a + b + c$

By comparing our derived equation $2x - y + 3z + 8 = 0$ with the given form $ax + by + cz + 8 = 0$, we can identify the coefficients:

• $a = 2$
• $b = -1$
• $c = 3$

Finally, we calculate the required value $a + b + c$: $a + b + c = 2 + (-1) + 3$ $a + b + c = 1 + 3$ $a + b + c = 4$

Why this step? This is the final step to answer the specific question posed by the problem.

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Common Mistakes & Tips

• Cross Product Calculation Errors: Be extremely careful with signs and the order of operations when calculating the determinant for the cross product. A single sign error can lead to an incorrect normal vector.
• Matching Equation Form: Always ensure the derived plane equation is adjusted to match the specified format (especially the constant term) before identifying coefficients $a, b, c$. Multiplying the entire equation by a non-zero scalar does not change the plane.
• Geometric Interpretation: Visualize the problem. A plane perpendicular to two others means its normal vector is perpendicular to their normal vectors. This geometric insight is key to understanding why the cross product is used.

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Summary

To find the equation of a plane passing through a given point and perpendicular to two other planes, the systematic approach involves first extracting the normal vectors of the two given planes. Their cross product then provides the normal vector of the required plane. Using this normal vector and the given point, the equation of the plane can be constructed using the point-normal form. Finally, the equation is simplified and adjusted to match the specified format, allowing for the identification of coefficients $a, b, c$ and the calculation of their sum.

The final answer is $\boxed{4}$, which corresponds to option (D).