1. Key Concepts and Formulas

This problem requires us to determine the equation of a plane that satisfies two specific conditions:

1. It passes through the line of intersection of two given planes.
2. It is perpendicular to a third given plane.

To achieve this, we will utilize the following fundamental concepts from 3D geometry:

• Equation of a Plane through the Line of Intersection of Two Planes: If $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$ are the Cartesian equations of two distinct 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 an arbitrary scalar constant. This equation represents a family of planes, and a specific value of $\lambda$ defines the unique plane that satisfies any additional given condition.

• Condition for Perpendicularity of Two Planes: Two planes, $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$, are perpendicular if and only if their normal vectors are orthogonal. The normal vector to a plane $Ax + By + Cz + D = 0$ is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. Thus, the condition for perpendicularity is: $\vec{n_1} \cdot \vec{n_2} = 0 \quad \Rightarrow \quad A_1A_2 + B_1B_2 + C_1C_2 = 0$

• Conversion between Cartesian and Vector Form of a Plane:

  • Cartesian to Vector: A plane given by $Ax + By + Cz + D = 0$ can be expressed in vector form as $\vec{r} \cdot (A\hat{i} + B\hat{j} + C\hat{k}) + D = 0$, where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is the position vector of any point on the plane. The vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ is the normal vector to the plane.
  • Vector to Cartesian: Conversely, a plane given by $\vec{r} \cdot \vec{n} + D = 0$ can be written as $Ax + By + Cz + D = 0$ by substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

2. Step-by-Step Solution

Step 2.1: Formulate the general equation of the required plane We are given two planes whose line of intersection defines the family of planes our desired plane belongs to. First, we write their equations in the standard $Ax+By+Cz+D=0$ form:

• Plane $P_1$: $x + y + z - 1 = 0$
• Plane $P_2$: $2x + 3y + 4z - 5 = 0$

Using the formula $P_1 + \lambda P_2 = 0$, the equation of any plane passing through the line of intersection of $P_1$ and $P_2$ is: $(x + y + z - 1) + \lambda (2x + 3y + 4z - 5) = 0$ To clearly identify the coefficients of $x, y, z$ and the constant term, we group them: $(1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z - (1 + 5\lambda) = 0 \quad \ldots (1)$ This is the Cartesian equation of the general plane. Our next step is to find the specific value of $\lambda$ that satisfies the second condition (perpendicularity).

Step 2.2: Identify normal vectors The normal vector to a plane $Ax + By + Cz + D = 0$ is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.

• Normal vector to plane (1): From equation (1), the coefficients of $x, y, z$ are $A_1 = (1 + 2\lambda)$, $B_1 = (1 + 3\lambda)$, and $C_1 = (1 + 4\lambda)$. So, the normal vector to our required plane is: $\vec{n_1} = (1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}$

• Normal vector to the perpendicular plane: The required plane must be perpendicular to the plane $P_3: x - y + z = 0$. The coefficients of $x, y, z$ for $P_3$ are $A_2 = 1$, $B_2 = -1$, and $C_2 = 1$. So, the normal vector to $P_3$ is: $\vec{n_2} = 1\hat{i} - 1\hat{j} + 1\hat{k} = \hat{i} - \hat{j} + \hat{k}$

Step 2.3: Apply the condition for perpendicularity to find $\lambda$ Since plane (1) is perpendicular to $P_3$, their normal vectors $\vec{n_1}$ and $\vec{n_2}$ must be orthogonal. This means their dot product must be zero: $\vec{n_1} \cdot \vec{n_2} = 0$ $((1 + 2\lambda)\hat{i} + (1 + 3\lambda)\hat{j} + (1 + 4\lambda)\hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = 0$ Performing the dot product: $(1 + 2\lambda)(1) + (1 + 3\lambda)(-1) + (1 + 4\lambda)(1) = 0$ $1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$ Combine the like terms: $(2\lambda - 3\lambda + 4\lambda) + (1 - 1 + 1) = 0$ $3\lambda + 1 = 0$ Solving for $\lambda$: $3\lambda = -1 \quad \Rightarrow \quad \lambda = -\frac{1}{3}$

Step 2.4: Determine the specific plane equation Now we substitute the value of $\lambda = -\frac{1}{3}$ back into the general equation of the plane (1): $x\left(1 + 2\left(-\frac{1}{3}\right)\right) + y\left(1 + 3\left(-\frac{1}{3}\right)\right) + z\left(1 + 4\left(-\frac{1}{3}\right)\right) - \left(1 + 5\left(-\frac{1}{3}\right)\right) = 0$ Simplify the coefficients: $x\left(1 - \frac{2}{3}\right) + y\left(1 - 1\right) + z\left(1 - \frac{4}{3}\right) - \left(1 - \frac{5}{3}\right) = 0$ $x\left(\frac{1}{3}\right) + y(0) + z\left(-\frac{1}{3}\right) - \left(-\frac{2}{3}\right) = 0$ $\frac{x}{3} - \frac{z}{3} + \frac{2}{3} = 0$ To simplify and remove the denominators, we multiply the entire equation by 3: $x - z + 2 = 0$ This is the Cartesian equation of the required plane.

Step 2.5: Convert to vector form and match options The Cartesian equation of the plane is $x - z + 2 = 0$. To convert this to the vector form $\vec{r} \cdot \vec{n} + D = 0$, we identify the normal vector $\vec{n}$ and the constant $D$. Here, the coefficients are $A=1$, $B=0$, $C=-1$, and the constant term is $D=2$. So, the normal vector is $\vec{n} = 1\hat{i} + 0\hat{j} - 1\hat{k} = \hat{i} - \hat{k}$. The vector equation of the plane is: $\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$ Comparing this result with the given options, this equation directly matches option (C).

3. Common Mistakes & Tips

• Operator Confusion (Dot vs. Cross Product): A common mistake is to confuse the dot product ($\vec{r} \cdot \vec{n} = d$) which defines a plane, with a cross product ($\vec{r} \times \vec{n} = \vec{c}$) which defines a line. Always remember that the equation of a plane involves a dot product with its normal vector.
• Sign Errors with Constant Terms: Be careful when writing planes as $Ax+By+Cz+D=0$. If the original plane is $Ax+By+Cz=d$, then $D=-d$. Consistent handling of signs is crucial, especially when substituting $\lambda$ into the constant part of the equation $P_1 + \lambda P_2 = 0$.
• Algebraic Precision: Mistakes often occur in simplifying expressions with $\lambda$ or when substituting its value back into the equation. Double-check all calculations, especially when dealing with fractions.

4. Summary

To find the equation of a plane passing through the line of intersection of two planes and perpendicular to a third plane, we first form the general equation of the plane $P_1 + \lambda P_2 = 0$. Then, we identify the normal vector of this general plane and the normal vector of the plane it needs to be perpendicular to. By applying the perpendicularity condition ($\vec{n_1} \cdot \vec{n_2} = 0$), we solve for the scalar $\lambda$. Substituting this value of $\lambda$ back into the general equation yields the specific Cartesian equation of the required plane. Finally, this Cartesian equation is converted into its vector form $\vec{r} \cdot \vec{n} + D = 0$. Following these steps, the derived vector equation for the plane is $\vec{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$.

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