1. Key Concepts and Formulas

To solve this problem, we will utilize two fundamental concepts from 3D geometry:

• Equation of a Plane Containing the Line of Intersection of Two Planes: If we have two planes, $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$, then the equation of any plane that passes through their line of intersection is given by the linear combination: $P_1 + \lambda P_2 = 0$ where $\lambda$ (lambda) is a scalar constant. This equation represents a family of planes, and we need to find the specific value of $\lambda$ that satisfies the given conditions.

• Condition for Parallel Planes: Two planes are parallel if and only if their normal vectors are parallel. For a plane $Ax+By+Cz+D=0$, its normal vector is $\vec{n} = \langle A, B, C \rangle$. If two planes have normal vectors $\vec{n}_1 = \langle A_1, B_1, C_1 \rangle$ and $\vec{n}_2 = \langle A_2, B_2, C_2 \rangle$, they are parallel if their direction ratios are proportional: $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$

2. Step-by-Step Solution

Our goal is to find the equation of a plane that satisfies two conditions: it contains a given line (which is the intersection of two other planes) and it is parallel to a third given plane.

Step 1: Formulate the Equation of the Family of Planes

The line is given by the intersection of two planes:

• Plane 1 ($P_1$): $2x - 5y + z = 3 \quad \Rightarrow \quad 2x - 5y + z - 3 = 0$
• Plane 2 ($P_2$): $x + y + 4z = 5 \quad \Rightarrow \quad x + y + 4z - 5 = 0$

According to the first key concept, any plane containing the line of intersection of $P_1$ and $P_2$ can be written as $P_1 + \lambda P_2 = 0$. Substituting the equations of $P_1$ and $P_2$: $(2x - 5y + z - 3) + \lambda (x + y + 4z - 5) = 0$ This is the general equation for the family of planes that pass through the given line.

Step 2: Express the Family of Planes in Standard Form and Identify its Normal Vector

To identify the normal vector and apply the parallelism condition, we need to rearrange the equation from Step 1 into the standard form $Ax + By + Cz + D = 0$. We do this by grouping the terms involving $x$, $y$, and $z$: $(2x + \lambda x) + (-5y + \lambda y) + (z + 4\lambda z) + (-3 - 5\lambda) = 0$ Factoring out $x, y, z$: $(2 + \lambda)x + (-5 + \lambda)y + (1 + 4\lambda)z + (-3 - 5\lambda) = 0 \quad \text{... (i)}$ This is the general equation of a plane from our family. The normal vector to this plane, $\vec{n}_1$, is given by the coefficients of $x, y, z$: $\vec{n}_1 = \langle (2 + \lambda), (-5 + \lambda), (1 + 4\lambda) \rangle$

Step 3: Apply the Condition for Parallel Planes

We are given that the required plane (equation (i)) is parallel to the plane $P_3: x + 3y + 6z = 1$. The normal vector to $P_3$ is $\vec{n}_2 = \langle 1, 3, 6 \rangle$.

For two planes to be parallel, their normal vectors must be parallel. This means their corresponding components must be proportional, as stated in our second key concept. Therefore, for plane (i) and plane $P_3$ to be parallel, we must have: $\frac{2 + \lambda}{1} = \frac{-5 + \lambda}{3} = \frac{1 + 4\lambda}{6}$

Step 4: Solve for the Scalar $\lambda$

We can find the value of $\lambda$ by equating any two of the ratios from Step 3. Let's use the first two ratios: $\frac{2 + \lambda}{1} = \frac{-5 + \lambda}{3}$ Cross-multiplying to solve for $\lambda$: $3(2 + \lambda) = 1(-5 + \lambda)$ $6 + 3\lambda = -5 + \lambda$ $3\lambda - \lambda = -5 - 6$ $2\lambda = -11$ $\lambda = -\frac{11}{2}$ (Self-check: We can verify this value using another pair of ratios, e.g., $\frac{-5 + \lambda}{3} = \frac{1 + 4\lambda}{6}$. Substituting $\lambda = -\frac{11}{2}$ into this gives $\frac{-5 - 11/2}{3} = \frac{1 + 4(-11/2)}{6} \Rightarrow \frac{-21/2}{3} = \frac{1 - 22}{6} \Rightarrow -\frac{7}{2} = -\frac{21}{6} \Rightarrow -\frac{7}{2} = -\frac{7}{2}$, confirming our value of $\lambda$.)

Step 5: Substitute $\lambda$ Back to Find the Specific Plane Equation

Now that we have $\lambda = -\frac{11}{2}$, we substitute this value back into the general equation of the family of planes (equation (i)): $(2 + \lambda)x + (-5 + \lambda)y + (1 + 4\lambda)z + (-3 - 5\lambda) = 0$ Substituting $\lambda = -\frac{11}{2}$: $\left(2 - \frac{11}{2}\right)x + \left(-5 - \frac{11}{2}\right)y + \left(1 + 4\left(-\frac{11}{2}\right)\right)z + \left(-3 - 5\left(-\frac{11}{2}\right)\right) = 0$ Let's calculate each coefficient:

• Coefficient of $x$: $2 - \frac{11}{2} = \frac{4 - 11}{2} = -\frac{7}{2}$
• Coefficient of $y$: $-5 - \frac{11}{2} = \frac{-10 - 11}{2} = -\frac{21}{2}$
• Coefficient of $z$: $1 + 4\left(-\frac{11}{2}\right) = 1 - 22 = -21$
• Constant term: $-3 - 5\left(-\frac{11}{2}\right) = -3 + \frac{55}{2} = \frac{-6 + 55}{2} = \frac{49}{2}$

Substituting these simplified coefficients back into the equation: $-\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2} = 0$ To clear the fractions and simplify, multiply the entire equation by $-2$: $(-2) \times \left(-\frac{7}{2}x - \frac{21}{2}y - 21z + \frac{49}{2}\right) = (-2) \times 0$ $7x + 21y + 42z - 49 = 0$ Finally, divide the entire equation by $7$ to obtain the simplest form: $x + 3y + 6z - 7 = 0$ $x + 3y + 6z = 7$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when rearranging plane equations into $Ax+By+Cz+D=0$ form and when substituting negative values for $\lambda$.
• Algebraic Precision: Mistakes often arise from incorrect fraction arithmetic or distribution. Double-check your calculations, especially when solving for $\lambda$.
• Normal Vector Identification: Ensure you correctly identify the coefficients of $x, y, z$ as the components of the normal vector.
• Simplification: Always simplify the final plane equation by dividing by any common factors to match the options provided.

4. Summary

We systematically found the equation of the plane by first expressing it as a family of planes passing through the intersection of the two given planes, using the form $P_1 + \lambda P_2 = 0$. Then, we used the condition that this plane must be parallel to a third given plane, which implies their normal vectors are proportional. This allowed us to solve for the scalar $\lambda$. Finally, substituting the value of $\lambda$ back into the family equation yielded the specific equation of the required plane, which is $x + 3y + 6z = 7$.

5. Final Answer

The equation of the plane is $x + 3y + 6z = 7$. The final answer is \boxed{x+3y+6z=7} which corresponds to option (A).