Key Concepts and Formulas

• Equation of a Plane through the Origin: A plane passing through the origin $(0,0,0)$ has an equation of the form $Ax + By + Cz = 0$, where $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ is its normal vector.
• Line Contained in a Plane: If a straight line with direction vector $\vec{d}$ lies within a plane, then the direction vector $\vec{d}$ must be perpendicular to the plane's normal vector $\vec{n}$. Mathematically, this means their dot product is zero: $\vec{d} \cdot \vec{n} = 0$.
• Normal Vector of a Plane Containing Two Lines: If a plane contains two non-parallel lines with direction vectors $\vec{d_a}$ and $\vec{d_b}$, then the normal vector to the plane can be found by taking the cross product of these direction vectors: $\vec{n} = \vec{d_a} \times \vec{d_b}$.
• Perpendicular Planes: If two planes are perpendicular to each other, their respective normal vectors are also perpendicular. If $\vec{n_1}$ is the normal vector to the first plane and $\vec{n_2}$ is the normal vector to the second plane, then $\vec{n_1} \cdot \vec{n_2} = 0$.

Step-by-Step Solution

Step 1: Determine conditions for the normal vector of the required plane (Plane P1). Let the equation of the required plane (Plane P1) be $Ax + By + Cz + D = 0$, and its normal vector be $\vec{n_1} = A\hat{i} + B\hat{j} + C\hat{k}$. The problem states that Plane P1 contains the straight line L1: ${x \over 2} = {y \over 3} = {z \over 4}$ The direction vector of L1 is $\vec{d_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$. Since L1 lies in P1, its direction vector $\vec{d_1}$ must be perpendicular to the normal vector $\vec{n_1}$. This gives us the first condition: $\vec{n_1} \cdot \vec{d_1} = 0$ $(A\hat{i} + B\hat{j} + C\hat{k}) \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = 0$ $2A + 3B + 4C = 0 \quad \text{(Equation 1)}$ Additionally, the line L1 passes through the origin $(0,0,0)$. Since P1 contains L1, P1 must also pass through the origin. Therefore, the constant term $D$ in the plane equation must be zero, simplifying the equation of P1 to $Ax + By + Cz = 0$.

Step 2: Find the normal vector of the second plane (Plane P2). The problem states that Plane P1 is perpendicular to another plane, let's call it Plane P2. Plane P2 contains two straight lines: L2: ${x \over 3} = {y \over 4} = {z \over 2}$ L3: ${x \over 4} = {y \over 2} = {z \over 3}$ The direction vector of L2 is $\vec{d_2} = 3\hat{i} + 4\hat{j} + 2\hat{k}$. The direction vector of L3 is $\vec{d_3} = 4\hat{i} + 2\hat{j} + 3\hat{k}$. Since both L2 and L3 lie in Plane P2, their direction vectors $\vec{d_2}$ and $\vec{d_3}$ are parallel to Plane P2. Consequently, the normal vector to Plane P2, $\vec{n_2}$, must be perpendicular to both $\vec{d_2}$ and $\vec{d_3}$. We can find $\vec{n_2}$ by taking the cross product of $\vec{d_2}$ and $\vec{d_3}$: $\vec{n_2} = \vec{d_2} \times \vec{d_3}$ $\vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{vmatrix}$ $\vec{n_2} = \hat{i}((4)(3) - (2)(2)) - \hat{j}((3)(3) - (2)(4)) + \hat{k}((3)(2) - (4)(4))$ $\vec{n_2} = \hat{i}(12 - 4) - \hat{j}(9 - 8) + \hat{k}(6 - 16)$ $\vec{n_2} = 8\hat{i} - \hat{j} - 10\hat{k}$ So, the direction ratios of the normal vector to Plane P2 are $(8, -1, -10)$.

Step 3: Apply the perpendicularity condition between Plane P1 and Plane P2. We are given that Plane P1 is perpendicular to Plane P2. This implies that their normal vectors, $\vec{n_1}$ and $\vec{n_2}$, are perpendicular. Therefore, their dot product must be zero: $\vec{n_1} \cdot \vec{n_2} = 0$ $(A\hat{i} + B\hat{j} + C\hat{k}) \cdot (8\hat{i} - \hat{j} - 10\hat{k}) = 0$ $8A - B - 10C = 0 \quad \text{(Equation 2)}$

Step 4: Solve the system of equations to find the direction ratios of $\vec{n_1}$. We now have a system of two linear equations for A, B, and C:

1. $2A + 3B + 4C = 0$
2. $8A - B - 10C = 0$

From Equation 2, we can express B in terms of A and C: $B = 8A - 10C$.

Substitute this expression for B into Equation 1: $2A + 3(8A - 10C) + 4C = 0$ $2A + 24A - 30C + 4C = 0$ $26A - 26C = 0$ $26A = 26C \implies A = C$.

Now substitute $A = C$ back into the expression for B: $B = 8A - 10(A)$ $B = 8A - 10A$ $B = -2A$.

Thus, the direction ratios of $\vec{n_1}$ are proportional to $(A, B, C) = (A, -2A, A)$. We can choose any non-zero value for A. For simplicity, let $A = 1$. Then, $A = 1$, $B = -2$, and $C = 1$. So, the normal vector to Plane P1 is $\vec{n_1} = \hat{i} - 2\hat{j} + \hat{k}$.

Step 5: Write the final equation of Plane P1. Using the direction ratios $(A, B, C) = (1, -2, 1)$ and the fact that the plane passes through the origin $(0,0,0)$, the equation of Plane P1 is: $Ax + By + Cz = 0$ $1x - 2y + 1z = 0$ $x - 2y + z = 0$

Common Mistakes & Tips

• Algebraic Errors in Cross Product: Carefully calculate the determinant for the cross product. A single sign error or arithmetic mistake will propagate through the solution.
• Incorrect Interpretation of Line in Plane: Remember that if a line lies in a plane, its direction vector is perpendicular to the plane's normal, AND any point on the line must satisfy the plane's equation. Using the origin $(0,0,0)$ from the first line simplifies the general plane equation from $Ax+By+Cz+D=0$ to $Ax+By+Cz=0$.
• Confusing Perpendicularity with Parallelism: For perpendicular planes, their normal vectors are perpendicular (dot product is zero). For parallel planes, their normal vectors are parallel (cross product is zero or components are proportional).

Summary We began by noting that the required plane contains the first given line, which passes through the origin. This allowed us to formulate a preliminary equation for the plane and its normal vector's components. Next, we determined the normal vector of the second plane by taking the cross product of the direction vectors of the two lines it contains. Finally, by applying the condition that the two planes are perpendicular (meaning their normal vectors are orthogonal), we formed a system of linear equations. Solving this system provided the direction ratios of the normal vector for the desired plane, leading directly to its equation.

Final Answer The equation of the plane is $x - 2y + z = 0$. The final answer is \boxed{x - 2y + z = 0}, which corresponds to option (A).