Here's a clear, educational, and well-structured solution to the problem:

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1. Key Concepts and Formulas

• Equation of a Plane Passing Through the Intersection of Two Planes: If the equations of two planes are given in vector form as $\overrightarrow r \cdot \overrightarrow{n_1} = d_1$ and $\overrightarrow r \cdot \overrightarrow{n_2} = d_2$, then the equation of any plane passing through their line of intersection is given by: $(\overrightarrow r \cdot \overrightarrow{n_1} - d_1) + \lambda (\overrightarrow r \cdot \overrightarrow{n_2} - d_2) = 0$ Here, $\lambda$ is a scalar parameter (an arbitrary constant). This equation represents a family of planes, all sharing the same line of intersection. A specific value of $\lambda$ defines a unique plane in this family.
• Vector Form of a Plane: The general vector equation of a plane is $\overrightarrow r \cdot \overrightarrow n = d$, where $\overrightarrow r$ is the position vector of any point on the plane, $\overrightarrow n$ is a normal vector to the plane, and $d$ is a constant.
• Dot Product: For two vectors $\overrightarrow A = A_x\widehat i + A_y\widehat j + A_z\widehat k$ and $\overrightarrow B = B_x\widehat i + B_y\widehat j + B_z\widehat k$, their dot product is $\overrightarrow A \cdot \overrightarrow B = A_x B_x + A_y B_y + A_z B_z$.

2. Step-by-Step Solution

Step 1: Identify the Given Planes and the Point We are given two planes and a point through which the required plane passes.

• Plane 1 ($P_1$): $\overrightarrow r \cdot (\widehat i + \widehat j + \widehat k) = 1$ To use the formula, we rewrite this as $\overrightarrow r \cdot (\widehat i + \widehat j + \widehat k) - 1 = 0$. Here, $\overrightarrow{n_1} = \widehat i + \widehat j + \widehat k$ and $d_1 = 1$.
• Plane 2 ($P_2$): $\overrightarrow r \cdot (\widehat i - 2\widehat j) = -2$ We rewrite this as $\overrightarrow r \cdot (\widehat i - 2\widehat j) + 2 = 0$. Here, $\overrightarrow{n_2} = \widehat i - 2\widehat j$ and $d_2 = -2$.
• The required plane passes through the point $(1, 0, 2)$. The position vector of this point is $\overrightarrow a = \widehat i + 0\widehat j + 2\widehat k = \widehat i + 2\widehat k$.

Step 2: Formulate the Equation of the Family of Planes Using the key concept for the equation of a plane passing through the intersection of two planes ($P_1 + \lambda P_2 = 0$), we write: $[\overrightarrow r \cdot (\widehat i + \widehat j + \widehat k) - 1] + \lambda [\overrightarrow r \cdot (\widehat i - 2\widehat j) + 2] = 0 \quad (*)$ Explanation: This equation represents all possible planes that contain the line formed by the intersection of $P_1$ and $P_2$. The parameter $\lambda$ helps us select the specific plane that also satisfies the additional condition (passing through the given point).

Step 3: Use the Given Point to Determine the Value of $\lambda$ Since the required plane passes through the point $(1, 0, 2)$, its position vector $\overrightarrow a = \widehat i + 2\widehat k$ must satisfy the equation of the plane. We substitute $\overrightarrow r = \widehat i + 2\widehat k$ into equation $(*)$: $[(\widehat i + 2\widehat k) \cdot (\widehat i + \widehat j + \widehat k) - 1] + \lambda [(\widehat i + 2\widehat k) \cdot (\widehat i - 2\widehat j) + 2] = 0$ Explanation: Any point lying on a plane must satisfy its equation. By substituting the coordinates of the given point, we can solve for the unique value of $\lambda$ that identifies our specific plane.

Step 4: Calculate Dot Products and Solve for $\lambda$ Let's evaluate the dot products:

• First term's dot product: $(\widehat i + 2\widehat k) \cdot (\widehat i + \widehat j + \widehat k) = (1)(1) + (0)(1) + (2)(1) = 1 + 0 + 2 = 3$
• Second term's dot product: $(\widehat i + 2\widehat k) \cdot (\widehat i - 2\widehat j) = (1)(1) + (0)(-2) + (2)(0) = 1 + 0 + 0 = 1$

Now, substitute these values back into the equation from Step 3: $[3 - 1] + \lambda [1 + 2] = 0$ $2 + \lambda (3) = 0$ $3\lambda = -2$ $\lambda = -\frac{2}{3}$ Explanation: We perform the scalar (dot) products carefully. The dot product of orthogonal unit vectors (e.g., $\widehat i \cdot \widehat j$) is zero, and that of identical unit vectors (e.g., $\widehat i \cdot \widehat i$) is one. After computing these, we solve the resulting linear equation for $\lambda$.

Step 5: Substitute the Value of $\lambda$ Back into the Family of Planes Equation Now that we have found $\lambda = -\frac{2}{3}$, we substitute it back into equation $(*)$ to get the specific equation of the plane: $[\overrightarrow r \cdot (\widehat i + \widehat j + \widehat k) - 1] + \left(-\frac{2}{3}\right) [\overrightarrow r \cdot (\widehat i - 2\widehat j) + 2] = 0$ Explanation: This step finalizes the equation for the unique plane that satisfies all given conditions.

Step 6: Simplify the Equation into Standard Vector Form To simplify and remove the fraction, multiply the entire equation by 3: $3[\overrightarrow r \cdot (\widehat i + \widehat j + \widehat k) - 1] - 2[\overrightarrow r \cdot (\widehat i - 2\widehat j) + 2] = 0$ Distribute the constants: $3(\overrightarrow r \cdot (\widehat i + \widehat j + \widehat k)) - 3 - 2(\overrightarrow r \cdot (\widehat i - 2\widehat j)) - 4 = 0$ Group the terms involving $\overrightarrow r$ and the constant terms: $\overrightarrow r \cdot [3(\widehat i + \widehat j + \widehat k) - 2(\widehat i - 2\widehat j)] - 3 - 4 = 0$ Simplify the normal vector inside the brackets: $3(\widehat i + \widehat j + \widehat k) - 2(\widehat i - 2\widehat j) = (3\widehat i + 3\widehat j + 3\widehat k) + (-2\widehat i + 4\widehat j)$ $= (3-2)\widehat i + (3+4)\widehat j + 3\widehat k$ $= \widehat i + 7\widehat j + 3\widehat k$ Simplify the constant term: $-3 - 4 = -7$ Substitute these back into the plane equation: $\overrightarrow r \cdot (\widehat i + 7\widehat j + 3\widehat k) - 7 = 0$ Finally, move the constant term to the right side to get the standard vector form: $\overrightarrow r \cdot (\widehat i + 7\widehat j + 3\widehat k) = 7$ Explanation: This involves algebraic manipulation. We use properties of dot products like $k(\overrightarrow r \cdot \overrightarrow A) = \overrightarrow r \cdot (k\overrightarrow A)$ and $\overrightarrow r \cdot \overrightarrow A + \overrightarrow r \cdot \overrightarrow B = \overrightarrow r \cdot (\overrightarrow A + \overrightarrow B)$ to combine the normal vectors and simplify the equation to the standard form $\overrightarrow r \cdot \overrightarrow n = d$.

3. Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when rewriting plane equations (e.g., $\overrightarrow r \cdot \overrightarrow n = d$ becomes $\overrightarrow r \cdot \overrightarrow n - d = 0$) and when distributing negative signs.
• Dot Product Accuracy: Ensure accurate calculation of dot products, especially when components are zero or negative. Remember $(\widehat i \cdot \widehat i = 1, \widehat i \cdot \widehat j = 0)$.
• Algebraic Simplification: Double-check distribution and combining like terms for both scalar and vector components. A small arithmetic error can lead to an incorrect final answer.

4. Summary

To find the equation of a plane passing through the intersection of two given planes and a specific point, we first form the general equation of the family of planes through their intersection, which involves a scalar parameter $\lambda$. Then, we use the coordinates of the given point to substitute into this general equation. This allows us to solve for the unique value of $\lambda$. Finally, we substitute this value of $\lambda$ back into the family equation and simplify it to obtain the required plane's equation in standard vector form. Following these steps, the calculation yields $\overrightarrow r \cdot (\widehat i + 7\widehat j + 3\widehat k) = 7$.

5. Final Answer

The vector equation of the plane is $\overrightarrow r \cdot (\widehat i + 7\widehat j + 3\widehat k) = 7$. Comparing this with the given options, it matches option (B).

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