Key Concepts and Formulas

1. Equation of a Plane Passing Through the Intersection of Two Planes: If a plane passes through the line of intersection of 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 its equation can be written as $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar constant. This equation represents a family of planes, all sharing the same line of intersection. A specific value of $\lambda$ is determined by an additional condition.

2. Finding a Point and Direction Vector of a Line of Intersection:

   • To find a point on the line formed by the intersection of two planes, set one coordinate (e.g., $x=0$) and solve the resulting system of two linear equations for the other two coordinates.
   • The direction vector of the line of intersection of two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$ is given by their cross product: $\vec{d} = \vec{n_1} \times \vec{n_2}$.

3. Equation of a Plane Containing Two Intersecting Lines: If two lines $L_1$ and $L_2$ intersect, the unique plane containing both of them can be found using their intersection point and their direction vectors. The normal vector of this plane is perpendicular to both direction vectors, i.e., $\vec{N} = \vec{d_1} \times \vec{d_2}$. The plane equation is $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$, where $(A,B,C)$ is the normal vector and $(x_0,y_0,z_0)$ is the intersection point.

4. Distance of a Point from a Plane: The distance $d$ of a point $(x_0, y_0, z_0)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula: $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Step-by-Step Solution

Understanding the Problem We need to find the equation of a plane that contains two given lines, $L_1$ and $L_2$. Each line is defined as the intersection of two other planes. Once we have the equation of this plane, we will calculate the distance from the origin $(0,0,0)$ to it.

Step 1: Formulating the Equation of the Plane Containing Line $L_1$

The required plane (let's denote it as $P_{req}$) contains line $L_1$. Line $L_1$ is the intersection of the planes: $P_1: 2x - 2y + 3z - 2 = 0$ $P_2: x - y + z + 1 = 0$

According to Key Concept 1, any plane passing through the line of intersection of $P_1$ and $P_2$ can be represented by the equation $P_1 + \lambda P_2 = 0$. So, the equation of the required plane $P_{req}$ is: $(2x - 2y + 3z - 2) + \lambda(x - y + z + 1) = 0 \quad (*)$ This can be rewritten as: $(2+\lambda)x + (-2-\lambda)y + (3+\lambda)z + (-2+\lambda) = 0$ Here, $\lambda$ is an unknown constant that we need to determine using the condition that $P_{req}$ also contains line $L_2$.

Why this step? This equation represents a family of planes, all containing $L_1$. To find the specific plane $P_{req}$ that also contains $L_2$, we need an additional condition. This condition will be that $P_{req}$ must contain every point on $L_2$. We will use one such point to find $\lambda$.

Step 2: Finding a Point on Line $L_2$

Line $L_2$ is the intersection of two other planes: $P_3: x + 2y - z - 3 = 0 \quad (1)$ $P_4: 3x - y + 2z - 1 = 0 \quad (2)$

To find a point on $L_2$, we can choose a convenient value for one of the coordinates and solve for the others. Let's set $x = 0$. Substituting $x = 0$ into equations (1) and (2) gives: From (1): $2y - z - 3 = 0 \Rightarrow 2y - z = 3 \quad (3)$ From (2): $-y + 2z - 1 = 0 \Rightarrow -y + 2z = 1 \quad (4)$

Multiply equation (4) by 2: $-2y + 4z = 2 \quad (5)$ Add equation (3) and (5): $(2y - z) + (-2y + 4z) = 3 + 2$ $3z = 5 \Rightarrow z = \frac{5}{3}$ Substitute $z = \frac{5}{3}$ into equation (3): $2y - \frac{5}{3} = 3$ $2y = 3 + \frac{5}{3} = \frac{9+5}{3} = \frac{14}{3}$ $y = \frac{7}{3}$ Thus, a point on line $L_2$ is $Q = \left(0, \frac{7}{3}, \frac{5}{3}\right)$.

Why this step? Since the required plane $P_{req}$ contains both $L_1$ and $L_2$, it must pass through every point on $L_2$. By finding just one point on $L_2$, we get a specific point that must lie on $P_{req}$. This point will serve as the condition to determine the unique value of $\lambda$.

Step 3: Determining the Value of $\lambda$

Since the plane $P_{req}$ contains line $L_2$, the point $Q\left(0, \frac{7}{3}, \frac{5}{3}\right)$ must satisfy the equation of $P_{req}$ from Step 1. Substitute $x=0$, $y=\frac{7}{3}$, $z=\frac{5}{3}$ into equation $(*)$: $\left(2(0) - 2\left(\frac{7}{3}\right) + 3\left(\frac{5}{3}\right) - 2\right) + \lambda\left(0 - \frac{7}{3} + \frac{5}{3} + 1\right) = 0$ $\left(0 - \frac{14}{3} + \frac{15}{3} - \frac{6}{3}\right) + \lambda\left(-\frac{2}{3} + \frac{3}{3}\right) = 0$ $\left(\frac{1}{3} - \frac{6}{3}\right) + \lambda\left(\frac{1}{3}\right) = 0$ $-\frac{5}{3} + \frac{\lambda}{3} = 0$ Multiplying by 3 to clear the denominators: $-5 + \lambda = 0 \Rightarrow \lambda = 5$

Why this step? This is a crucial step. By ensuring that $P_{req}$ contains a point from $L_2$, we fix the value of $\lambda$. This uniquely identifies the required plane from the infinite family of planes that contain $L_1$.

Step 4: Finding the Equation of the Required Plane ($P_{req}$)

Now that we have $\lambda = 5$, substitute this value back into the general equation of plane $P_{req}$ from Step 1: $(2x - 2y + 3z - 2) + 5(x - y + z + 1) = 0$ Expand and combine like terms: $2x - 2y + 3z - 2 + 5x - 5y + 5z + 5 = 0$ $(2+5)x + (-2-5)y + (3+5)z + (-2+5) = 0$ $7x - 7y + 8z + 3 = 0$ This is the equation of the plane $P_{req}$ that contains both lines $L_1$ and $L_2$.

Why this step? This provides the explicit algebraic representation of the plane, which is necessary for the final calculation of the distance from the origin.

Step 5: Calculating the Distance from the Origin

We need to find the distance of the origin $(x_0, y_0, z_0) = (0, 0, 0)$ from the plane $7x - 7y + 8z + 3 = 0$. Using the distance formula $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$: Here, $A=7$, $B=-7$, $C=8$, $D=3$.

$d = \frac{|7(0) - 7(0) + 8(0) + 3|}{\sqrt{7^2 + (-7)^2 + 8^2}}$ $d = \frac{|3|}{\sqrt{49 + 49 + 64}}$ $d = \frac{3}{\sqrt{162}}$ To simplify the square root, we find perfect square factors of 162: $\sqrt{162} = \sqrt{81 \times 2} = \sqrt{81} \times \sqrt{2} = 9\sqrt{2}$ So, the distance is: $d = \frac{3}{9\sqrt{2}}$ $d = \frac{1}{3\sqrt{2}}$

Why this step? This is the final objective of the problem. Applying the standard distance formula yields the required numerical answer.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with signs and arithmetic, especially when combining terms and solving for $\lambda$. A small calculation error can propagate and lead to an incorrect final answer.
• Simplifying Radicals: Always simplify square roots completely to match the format of the options provided.
• Verifying Line Intersection (Optional but good practice): The method assumes that a unique plane exists, which is true if the lines intersect (or are parallel and distinct). In this problem, the lines $L_1$ and $L_2$ intersect at $(-1, 4, 4)$, confirming the existence of a unique plane.

Summary

We first found the equation of the plane containing line $L_1$ in terms of an unknown parameter $\lambda$. Then, by finding a point on line $L_2$ and substituting its coordinates into the plane equation, we determined the value of $\lambda$. This allowed us to find the explicit equation of the plane containing both lines. Finally, we used the standard formula to calculate the distance from the origin to this plane. The distance was found to be $\frac{1}{3\sqrt{2}}$.

The final answer is $\boxed{\frac{1}{3\sqrt{2}}}$, which corresponds to option (C).