Key Concepts and Formulas

1. Equation of a Plane (Point-Normal Form): The equation of a plane passing through a point $(x_1, y_1, z_1)$ and having a normal vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ is given by: $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$ This form is fundamental for constructing a plane's equation when a point on it and its orientation (via the normal vector) are known.

2. Normal Vector from Plane Equation: For a plane given in the general form $Ax + By + Cz + D = 0$, the vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ is its normal vector. This vector is perpendicular to every vector lying in the plane.

3. Perpendicular Planes and Normal Vectors: If two planes are perpendicular to each other, their respective normal vectors are also perpendicular. This means the dot product of their normal vectors is zero. If a plane is perpendicular to two other planes, its normal vector must be perpendicular to the normal vectors of both those planes.

4. Cross Product of Vectors: The cross product of two non-parallel vectors, $\vec{a}$ and $\vec{b}$, results in a new vector $\vec{a} \times \vec{b}$ that is perpendicular to both $\vec{a}$ and $\vec{b}$. This property is crucial for finding a vector that is simultaneously orthogonal to two given vectors.

Step-by-Step Solution

Step 1: Identify Given Information We need to find the equation of a plane that satisfies two conditions:

• It passes through a specific point: $P(1, 2, -3)$. This will be our $(x_1, y_1, z_1)$.
• It is perpendicular to two other given planes:
  • Plane 1 ($P_1$): $3x + y - 2z = 5$
  • Plane 2 ($P_2$): $2x - 5y - z = 7$

Step 2: Extract Normal Vectors of the Given Planes The normal vector of a plane $Ax + By + Cz + D = 0$ is simply $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. We need these normal vectors to understand the orientation of $P_1$ and $P_2$.

• For Plane 1 ($P_1: 3x + y - 2z = 5$), the normal vector is $\vec{n_1} = 3\hat{i} + 1\hat{j} - 2\hat{k}$.
• For Plane 2 ($P_2: 2x - 5y - z = 7$), the normal vector is $\vec{n_2} = 2\hat{i} - 5\hat{j} - 1\hat{k}$.

Step 3: Determine the Normal Vector of the Required Plane Let the required plane be $P$. Since $P$ is perpendicular to both $P_1$ and $P_2$, its normal vector, say $\vec{n}$, must be perpendicular to both $\vec{n_1}$ and $\vec{n_2}$. The cross product of two vectors yields a vector that is perpendicular to both original vectors. Therefore, we can find the normal vector $\vec{n}$ of our required plane $P$ by taking the cross product of $\vec{n_1}$ and $\vec{n_2}$. $\vec{n} = \vec{n_1} \times \vec{n_2}$ We calculate the cross product using the determinant form: $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1 \end{vmatrix}$ Expanding the determinant along the first row: $\vec{n} = \hat{i}((1)(-1) - (-2)(-5)) - \hat{j}((3)(-1) - (-2)(2)) + \hat{k}((3)(-5) - (1)(2))$ $\vec{n} = \hat{i}(-1 - 10) - \hat{j}(-3 - (-4)) + \hat{k}(-15 - 2)$ $\vec{n} = \hat{i}(-11) - \hat{j}(-3 + 4) + \hat{k}(-17)$ $\vec{n} = -11\hat{i} - \hat{j} - 17\hat{k}$ This vector $\vec{n}$ is the normal vector for our required plane. Any non-zero scalar multiple of $\vec{n}$ would also be a valid normal vector for the same plane (e.g., multiplying by $-1$ to get $11\hat{i} + \hat{j} + 17\hat{k}$).

Step 4: Formulate the Equation of the Required Plane We now have the normal vector $\vec{n} = -11\hat{i} - \hat{j} - 17\hat{k}$ (so $A = -11, B = -1, C = -17$) and the point $P(1, 2, -3)$ (so $x_1 = 1, y_1 = 2, z_1 = -3$) through which the plane passes.

Using the point-normal form of the plane equation: $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$ Substitute the values: $-11(x - 1) + (-1)(y - 2) + (-17)(z - (-3)) = 0$ $-11(x - 1) - 1(y - 2) - 17(z + 3) = 0$ Distribute the coefficients: $-11x + 11 - y + 2 - 17z - 51 = 0$ Combine the constant terms: $-11x - y - 17z + (11 + 2 - 51) = 0$ $-11x - y - 17z - 38 = 0$ To make the leading coefficient positive (a common convention), we can multiply the entire equation by $-1$: $11x + y + 17z + 38 = 0$

Common Mistakes & Tips

• Sign Errors: Be meticulous with signs throughout the calculation, especially during the cross product expansion and substitution into the point-normal form. A single sign error can lead to an incorrect final equation.
• Order of Cross Product: Remember that $\vec{n_1} \times \vec{n_2}$ and $\vec{n_2} \times \vec{n_1}$ result in vectors that are opposite in direction. Both are valid normal vectors for the same plane, differing only by a scalar multiple of -1, and will lead to equivalent plane equations (e.g., $Ax + By + Cz + D = 0$ is equivalent to $-Ax - By - Cz - D = 0$).
• Verification Step: After finding the equation of the plane, it's good practice to verify the conditions.
  1. Check if the given point $(1, 2, -3)$ satisfies the equation: $11(1) + (2) + 17(-3) + 38 = 11 + 2 - 51 + 38 = 51 - 51 = 0$. (The point lies on the plane).
  2. Check if the plane's normal vector $\vec{n} = 11\hat{i} + \hat{j} + 17\hat{k}$ is perpendicular to $\vec{n_1}$ and $\vec{n_2}$:
     • $\vec{n} \cdot \vec{n_1} = (11)(3) + (1)(1) + (17)(-2) = 33 + 1 - 34 = 0$. (Perpendicular to $P_1$).
     • $\vec{n} \cdot \vec{n_2} = (11)(2) + (1)(-5) + (17)(-1) = 22 - 5 - 17 = 0$. (Perpendicular to $P_2$). Since both conditions are met, our derived equation is correct.

Summary

To find the equation of a plane that passes through a given point and is perpendicular to two other planes, the key strategy is to first determine its normal vector. This is achieved by taking the cross product of the normal vectors of the two planes to which it is perpendicular. Once the normal vector is found, along with the given point on the plane, the point-normal form of the plane equation can be used to derive the final equation. Careful calculation of the cross product and substitution is essential.

The final answer is $\boxed{\text{11x + y + 17z + 38 = 0}}$, which corresponds to option (D).