Key Concepts and Formulas

To solve this problem, we'll utilize several fundamental concepts from 3D Geometry:

• Equation of a Plane Containing the Line of Intersection of Two Planes: If two planes are given by $P_1: A_1x+B_1y+C_1z+D_1=0$ and $P_2: A_2x+B_2y+C_2z+D_2=0$, then any plane containing their line of intersection can be represented by the equation $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar constant.
• Normal Vector of a Plane:
  • For a plane in Cartesian form $Ax+By+Cz+D=0$, its normal vector is $\vec{n} = A\hat{i}+B\hat{j}+C\hat{k}$.
  • For a plane in parametric vector form $\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}$, where $\vec{b}$ and $\vec{c}$ are two non-parallel vectors lying in the plane, its normal vector can be found by their cross product: $\vec{n} = \vec{b} \times \vec{c}$.
• Condition for Perpendicular Planes: Two planes are perpendicular if and only if their normal vectors are orthogonal (perpendicular). Mathematically, if $\vec{n_1}$ and $\vec{n_2}$ are the normal vectors of the two planes, then their dot product must be zero: $\vec{n_1} \cdot \vec{n_2} = 0$.

Step-by-Step Solution

Step 1: Formulate the Equation of the Required Plane (P)

The problem states that the required plane, let's call it $P$, contains the line of intersection of two given planes: $P_1: x+2y+3z-4=0$ $P_2: 2x+y-z+5=0$

According to the key concept, the equation of plane $P$ can be written as $P_1 + \lambda P_2 = 0$: $(x+2y+3z-4) + \lambda(2x+y-z+5) = 0$ To identify the normal vector of this plane, we rearrange the equation into the standard Cartesian form $Ax+By+Cz+D=0$: $(1+2\lambda)x + (2+\lambda)y + (3-\lambda)z + (-4+5\lambda) = 0$ The normal vector of plane $P$, denoted as $\vec{n_P}$, consists of the coefficients of $x, y, z$: $\vec{n_P} = (1+2\lambda)\hat{i} + (2+\lambda)\hat{j} + (3-\lambda)\hat{k}$

Step 2: Determine the Normal Vector of the Second Given Plane (P')

The second given plane, $P'$, is provided in parametric vector form: $\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$ In this form, $\vec{a} = \hat{i}-\hat{j}$ is a position vector of a point on the plane, and $\vec{b} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{c} = \hat{i}-2 \hat{j}+3 \hat{k}$ are two direction vectors that lie within the plane.

The normal vector $\vec{n_{P'}}$ to this plane is perpendicular to both $\vec{b}$ and $\vec{c}$, and thus can be found by their cross product: $\vec{n_{P'}} = \vec{b} \times \vec{c} = (\hat{i}+\hat{j}+\hat{k}) \times (\hat{i}-2 \hat{j}+3 \hat{k})$ We calculate the cross product using a determinant: $\vec{n_{P'}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix}$ $\vec{n_{P'}} = \hat{i}((1)(3) - (1)(-2)) - \hat{j}((1)(3) - (1)(1)) + \hat{k}((1)(-2) - (1)(1))$ $\vec{n_{P'}} = \hat{i}(3 + 2) - \hat{j}(3 - 1) + \hat{k}(-2 - 1)$ $\vec{n_{P'}} = 5\hat{i} - 2\hat{j} - 3\hat{k}$

Step 3: Apply the Perpendicularity Condition to Find $\lambda$

We are given that plane $P$ is perpendicular to plane $P'$. This means their normal vectors, $\vec{n_P}$ and $\vec{n_{P'}}$, must be orthogonal. The dot product of orthogonal vectors is zero: $\vec{n_P} \cdot \vec{n_{P'}} = 0$ Substitute the expressions for $\vec{n_P}$ and $\vec{n_{P'}}$: $((1+2\lambda)\hat{i} + (2+\lambda)\hat{j} + (3-\lambda)\hat{k}) \cdot (5\hat{i} - 2\hat{j} - 3\hat{k}) = 0$ Perform the dot product: $5(1+2\lambda) - 2(2+\lambda) - 3(3-\lambda) = 0$ Now, solve this linear equation for $\lambda$: $5 + 10\lambda - 4 - 2\lambda - 9 + 3\lambda = 0$ Combine like terms: $(5 - 4 - 9) + (10\lambda - 2\lambda + 3\lambda) = 0$ $-8 + 11\lambda = 0$ $11\lambda = 8$ $\lambda = \frac{8}{11}$

Step 4: Substitute $\lambda$ to Obtain the Final Plane Equation

Substitute the value of $\lambda = \frac{8}{11}$ back into the equation of plane $P$ from Step 1: $(1+2\lambda)x + (2+\lambda)y + (3-\lambda)z + (-4+5\lambda) = 0$ Calculate the coefficients:

• Coefficient of $x$: $1 + 2\left(\frac{8}{11}\right) = 1 + \frac{16}{11} = \frac{11+16}{11} = \frac{27}{11}$
• Coefficient of $y$: $2 + \frac{8}{11} = \frac{22+8}{11} = \frac{30}{11}$
• Coefficient of $z$: $3 - \frac{8}{11} = \frac{33-8}{11} = \frac{25}{11}$
• Constant term: $-4 + 5\left(\frac{8}{11}\right) = -4 + \frac{40}{11} = \frac{-44+40}{11} = \frac{-4}{11}$

So the equation of plane $P$ is: $\frac{27}{11}x + \frac{30}{11}y + \frac{25}{11}z - \frac{4}{11} = 0$ To clear the denominators, multiply the entire equation by $11$: $27x + 30y + 25z - 4 = 0$ Rearrange this to match the given form $ax+by+cz=4$: $27x + 30y + 25z = 4$

Step 5: Calculate the Required Expression $(a-b+c)$

By comparing our derived equation $27x + 30y + 25z = 4$ with the general form $ax+by+cz=4$, we can identify the coefficients: $a = 27$ $b = 30$ $c = 25$

Now, substitute these values into the expression $(a-b+c)$: $a-b+c = 27 - 30 + 25$ $a-b+c = -3 + 25$ $a-b+c = 22$

Common Mistakes & Tips

• Arithmetic Errors: Solving for $\lambda$ involves several terms and fractions. Be meticulous with calculations, especially when combining terms or performing dot products. A small error can propagate and invalidate the final answer.
• Vector Operations: Ensure correct calculation of the cross product for the normal vector and the dot product for the perpendicularity condition. Remember the order of terms in the cross product determinant.
• Matching Equation Format: After finding the plane equation, ensure it matches the target form $ax+by+cz=4$. If your constant term is different (e.g., $27x+30y+25z=8$), you must divide or multiply the entire equation by a scalar to make the constant term equal to 4, which would then change $a, b, c$ accordingly. In this problem, it directly matched.

Summary

We first established the general equation of the plane containing the line of intersection of the two given planes in terms of an unknown parameter $\lambda$. Then, we found the normal vector of this plane. Next, we determined the normal vector of the second given plane from its parametric form using the cross product. By applying the condition that the two planes are perpendicular (their normal vectors' dot product is zero), we solved for $\lambda$. Finally, substituting the value of $\lambda$ back into the plane equation and matching it to the given form $ax+by+cz=4$, we identified $a, b, c$ and calculated the required expression $(a-b+c)$.

The final answer is $\boxed{\text{22}}$, which corresponds to option (B).