Key Concepts and Formulas

1. Equation of a Plane Passing 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 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 expression represents a family of planes, and a specific value of $\lambda$ will define the unique plane that satisfies additional given conditions.

2. Condition for a Plane to be Parallel to a Line: A plane with the equation $Ax + By + Cz + D = 0$ has a normal vector $\vec{n} = A\widehat{i} + B\widehat{j} + C\widehat{k}$. A line has a direction vector $\vec{d} = l\widehat{i} + m\widehat{j} + n\widehat{k}$. If the plane is parallel to the line, its normal vector $\vec{n}$ must be perpendicular to the line's direction vector $\vec{d}$. Mathematically, this means their dot product is zero: $\vec{n} \cdot \vec{d} = 0 \quad \text{or} \quad Al + Bm + Cn = 0$ For the x-axis, its direction vector is $\vec{d} = \widehat{i}$ (or direction ratios $(1, 0, 0)$).

3. Conversion Between Vector and Cartesian Forms of a Plane: The vector equation of a plane $\vec{r} \cdot \vec{n} = d$ can be written in Cartesian form as $Ax + By + Cz = d$, where $\vec{n} = A\widehat{i} + B\widehat{j} + C\widehat{k}$. Conversely, a Cartesian equation $Ax + By + Cz + D = 0$ can be written as $\vec{r} \cdot (A\widehat{i} + B\widehat{j} + C\widehat{k}) + D = 0$.

Step-by-Step Solution

Step 1: Convert the Given Plane Equations to Cartesian Form We begin by converting the given vector equations of the planes into their Cartesian (scalar) forms, as the formula for the line of intersection ($P_1 + \lambda P_2 = 0$) is typically applied with Cartesian equations. Recall that if $\vec{r} = x\widehat{i} + y\widehat{j} + z\widehat{k}$, then $\vec{r} \cdot (A\widehat{i} + B\widehat{j} + C\widehat{k}) = Ax + By + Cz$.

• Plane 1: The first plane is given by $\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$. Substituting $\vec{r} = x\widehat{i} + y\widehat{j} + z\widehat{k}$: $(x\widehat{i} + y\widehat{j} + z\widehat{k}) \cdot (\widehat{i} + \widehat{j} + \widehat{k}) = 1$ $x + y + z = 1$ To use it in the $P_1 + \lambda P_2 = 0$ form, we rearrange it so all terms are on one side: $P_1: x + y + z - 1 = 0$

• Plane 2: The second plane is given by $\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0$. Substituting $\vec{r} = x\widehat{i} + y\widehat{j} + z\widehat{k}$: $(x\widehat{i} + y\widehat{j} + z\widehat{k}) \cdot (2\widehat{i} + 3\widehat{j} - \widehat{k}) + 4 = 0$ $2x + 3y - z + 4 = 0$ This equation is already in the $P_2 = 0$ form: $P_2: 2x + 3y - z + 4 = 0$

Step 2: Formulate the Equation of the Family of Planes Passing Through the Line of Intersection Using the key concept, the equation of any plane passing through the line of intersection of $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$. We introduce $\lambda$ as a scalar constant, which we will determine using the additional condition. Substituting the Cartesian forms of $P_1$ and $P_2$: $(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0$

Step 3: Rearrange the Equation into Standard Cartesian Form To easily identify the coefficients of $x, y, z$ (which form the normal vector of the plane), we group the terms involving $x, y, z$ and the constant term. This prepares the equation for applying the parallelism condition. $x + y + z - 1 + 2\lambda x + 3\lambda y - \lambda z + 4\lambda = 0$ Grouping terms: $(1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0$ This is the general Cartesian equation of the required plane. Its normal vector is $\vec{n} = (1 + 2\lambda)\widehat{i} + (1 + 3\lambda)\widehat{j} + (1 - \lambda)\widehat{k}$.

Step 4: Apply the Condition for Parallelism to the x-axis The problem states that the required plane is parallel to the x-axis. According to our key concepts, if a plane is parallel to a line, its normal vector must be perpendicular to the direction vector of that line.

• The direction vector of the x-axis is $\vec{d} = \widehat{i}$ (or direction ratios $(1, 0, 0)$).
• The normal vector of our plane is $\vec{n} = (1 + 2\lambda)\widehat{i} + (1 + 3\lambda)\widehat{j} + (1 - \lambda)\widehat{k}$.

The condition for perpendicularity is $\vec{n} \cdot \vec{d} = 0$: $((1 + 2\lambda)\widehat{i} + (1 + 3\lambda)\widehat{j} + (1 - \lambda)\widehat{k}) \cdot (\widehat{i} + 0\widehat{j} + 0\widehat{k}) = 0$ This simplifies to: $(1 + 2\lambda)(1) + (1 + 3\lambda)(0) + (1 - \lambda)(0) = 0$ $1 + 2\lambda = 0$ Solving for $\lambda$: $2\lambda = -1$ $\lambda = -\frac{1}{2}$

Step 5: Substitute the Value of $\lambda$ Back into the Plane Equation Now that we have found the specific value of $\lambda = -1/2$, we substitute it back into the general equation of the family of planes from Step 3. This will give us the unique equation of the plane that satisfies both given conditions. $(1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0$ Substitute $\lambda = -1/2$: $(1 + 2(-\frac{1}{2}))x + (1 + 3(-\frac{1}{2}))y + (1 - (-\frac{1}{2}))z + (4(-\frac{1}{2}) - 1) = 0$ Calculate each coefficient:

• Coefficient of $x$: $1 - 1 = 0$
• Coefficient of $y$: $1 - \frac{3}{2} = -\frac{1}{2}$
• Coefficient of $z$: $1 + \frac{1}{2} = \frac{3}{2}$
• Constant term: $-2 - 1 = -3$

So, the equation of the required plane in Cartesian form is: $0x - \frac{1}{2}y + \frac{3}{2}z - 3 = 0$

Step 6: Simplify and Convert to Vector Form To simplify the equation and match the format of the options, we can multiply the entire equation by $-2$ to eliminate fractions and obtain integer coefficients: $(-2) \left( -\frac{1}{2}y + \frac{3}{2}z - 3 \right) = 0 \times (-2)$ $y - 3z + 6 = 0$ Finally, convert this Cartesian equation back to vector form. Recall that a plane $Ax + By + Cz + D = 0$ can be written as $\vec{r} \cdot (A\widehat{i} + B\widehat{j} + C\widehat{k}) + D = 0$. Here, $A=0$, $B=1$, $C=-3$, and $D=6$. Therefore, the vector equation of the plane is: $\overrightarrow r .\left( {0\widehat i + 1\widehat j - 3\widehat k} \right) + 6 = 0$ $\overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0$

Common Mistakes & Tips

• Sign Conventions: Be meticulous with signs when moving terms across the equality sign (e.g., $x+y+z=1$ becomes $x+y+z-1=0$) and when dealing with negative values for $\lambda$.
• Direction Ratios of Axes: Accurately identify the direction vector for the specified axis. For the x-axis, it's $\widehat{i}$ (or $(1,0,0)$); for the y-axis, it's $\widehat{j}$ (or $(0,1,0)$); and for the z-axis, it's $\widehat{k}$ (or $(0,0,1)$).
• Parallel vs. Perpendicular: Clearly distinguish the conditions: a plane parallel to a line means the plane's normal vector is perpendicular to the line's direction vector ($\vec{n} \cdot \vec{d} = 0$).

Summary

This problem is a quintessential example of combining two fundamental concepts in 3D geometry: finding the equation of a plane through the line of intersection of two given planes and applying a geometric condition (parallelism to an axis) to uniquely determine that plane. The solution involved converting the initial vector equations to Cartesian form, using the $P_1 + \lambda P_2 = 0$ formula to represent the family of planes, and then leveraging the parallelism condition (normal vector perpendicular to the x-axis direction vector) to solve for the scalar $\lambda$. Substituting this value back into the general equation and simplifying yielded the specific Cartesian equation, which was finally converted back to vector form to match the options.

The final answer is \boxed{\overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0}, which corresponds to option (A).