Key Concepts and Formulas

1. Line of Intersection of Two Planes:

   • Given two planes $P_1: \overrightarrow r \cdot \overrightarrow n_1 = d_1$ and $P_2: \overrightarrow r \cdot \overrightarrow n_2 = d_2$, the direction vector $\overrightarrow a$ of their line of intersection $L$ is perpendicular to both normal vectors $\overrightarrow n_1$ and $\overrightarrow n_2$. Thus, $\overrightarrow a = \overrightarrow n_1 \times \overrightarrow n_2$.
   • To find a point $Q(x_0, y_0, z_0)$ on $L$, substitute $x=0$ (or $y=0$ or $z=0$) into both Cartesian plane equations and solve the resulting system of two linear equations for the other two coordinates.
   • The equation of the line $L$ in symmetric form is $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \lambda$, where $\overrightarrow a = a\widehat i + b\widehat j + c\widehat k$.

2. Foot of Perpendicular from a Point to a Line:

   • Given an external point $A(x_1, y_1, z_1)$ and a line $L: \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \lambda$, any point $P(\alpha, \beta, \gamma)$ on $L$ can be expressed parametrically as $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$.
   • The vector $\overrightarrow{AP}$ connecting the external point $A$ to the foot of the perpendicular $P$ is perpendicular to the line $L$. Therefore, its dot product with the direction vector $\overrightarrow a$ of $L$ must be zero: $\overrightarrow{AP} \cdot \overrightarrow a = 0$.
   • Solving this equation for $\lambda$ gives the specific parameter value for the foot of the perpendicular. Substituting this $\lambda$ back into the parametric coordinates yields the coordinates of $P$.

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Step-by-Step Solution

Step 1: Convert Plane Equations to Cartesian Form We are given the vector equations of two planes: Plane 1 ($P_1$): $\overrightarrow r .(\widehat i - \widehat j + 2\widehat k) = 2$ Plane 2 ($P_2$): $\overrightarrow r .(2\widehat i + \widehat j - \widehat k) = 2$

Let $\overrightarrow r = x\widehat i + y\widehat j + z\widehat k$. Substituting this into the equations: For $P_1$: $(x\widehat i + y\widehat j + z\widehat k) \cdot (\widehat i - \widehat j + 2\widehat k) = 2$ $P_1: x - y + 2z = 2$ For $P_2$: $(x\widehat i + y\widehat j + z\widehat k) \cdot (2\widehat i + \widehat j - \widehat k) = 2$ $P_2: 2x + y - z = 2$ These are the Cartesian equations of the two planes.

Step 2: Find the Direction Vector of the Line of Intersection (L) The normal vector to $P_1$ is $\overrightarrow n_1 = \widehat i - \widehat j + 2\widehat k = \langle 1, -1, 2 \rangle$. The normal vector to $P_2$ is $\overrightarrow n_2 = 2\widehat i + \widehat j - \widehat k = \langle 2, 1, -1 \rangle$.

The direction vector $\overrightarrow a$ of the line of intersection $L$ is perpendicular to both $\overrightarrow n_1$ and $\overrightarrow n_2$. We find it using the cross product: $\overrightarrow a = \overrightarrow n_1 \times \overrightarrow n_2 = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{vmatrix}$ $\overrightarrow a = \widehat i ((-1)(-1) - (2)(1)) - \widehat j ((1)(-1) - (2)(2)) + \widehat k ((1)(1) - (-1)(2))$ $\overrightarrow a = \widehat i (1 - 2) - \widehat j (-1 - 4) + \widehat k (1 + 2)$ $\overrightarrow a = -\widehat i + 5\widehat j + 3\widehat k = \langle -1, 5, 3 \rangle$

Step 3: Find a Point on the Line of Intersection (L) To find a point on $L$, we can set one coordinate to zero. Let's set $z=0$ in the Cartesian equations from Step 1: From $P_1: x - y + 2(0) = 2 \Rightarrow x - y = 2 \quad \text{(Equation 1)}$ From $P_2: 2x + y - (0) = 2 \Rightarrow 2x + y = 2 \quad \text{(Equation 2)}$

Now, solve this system of linear equations: Add (Equation 1) and (Equation 2): $(x - y) + (2x + y) = 2 + 2$ $3x = 4 \Rightarrow x = \frac{4}{3}$

Substitute $x = \frac{4}{3}$ into (Equation 1): $\frac{4}{3} - y = 2$ $y = \frac{4}{3} - 2 = \frac{4 - 6}{3} = -\frac{2}{3}$

So, a point on the line of intersection is $Q\left(\frac{4}{3}, -\frac{2}{3}, 0\right)$.

Step 4: Write the Equation of the Line of Intersection (L) Using the point $Q\left(\frac{4}{3}, -\frac{2}{3}, 0\right)$ and the direction vector $\overrightarrow a = \langle -1, 5, 3 \rangle$, the symmetric form of the line $L$ is: $L: \frac{x - \frac{4}{3}}{-1} = \frac{y - (-\frac{2}{3})}{5} = \frac{z - 0}{3}$ Let this ratio be equal to a parameter $\lambda$: $\frac{x - \frac{4}{3}}{-1} = \frac{y + \frac{2}{3}}{5} = \frac{z}{3} = \lambda$ Thus, any point $P(\alpha, \beta, \gamma)$ on the line $L$ can be expressed parametrically as: $P(\alpha, \beta, \gamma) = \left(-\lambda + \frac{4}{3}, 5\lambda - \frac{2}{3}, 3\lambda\right)$

Step 5: Find the Foot of the Perpendicular P from Point $A(1, 2, 0)$ to Line L The given external point is $A(1, 2, 0)$. The foot of the perpendicular $P(\alpha, \beta, \gamma)$ is on $L$. Form the vector $\overrightarrow{AP}$: $\overrightarrow{AP} = \left\langle \left(-\lambda + \frac{4}{3} - 1\right), \left(5\lambda - \frac{2}{3} - 2\right), (3\lambda - 0) \right\rangle$ $\overrightarrow{AP} = \left\langle \left(-\lambda + \frac{1}{3}\right), \left(5\lambda - \frac{8}{3}\right), 3\lambda \right\rangle$ Since $\overrightarrow{AP}$ is perpendicular to the line $L$, it must be perpendicular to the direction vector $\overrightarrow a = \langle -1, 5, 3 \rangle$. Their dot product must be zero: $\overrightarrow{AP} \cdot \overrightarrow a = 0$ $\left(-\lambda + \frac{1}{3}\right)(-1) + \left(5\lambda - \frac{8}{3}\right)(5) + (3\lambda)(3) = 0$ $\left(\lambda - \frac{1}{3}\right) + \left(25\lambda - \frac{40}{3}\right) + 9\lambda = 0$ Combine terms with $\lambda$ and constant terms: $(1 + 25 + 9)\lambda + \left(-\frac{1}{3} - \frac{40}{3}\right) = 0$ $35\lambda - \frac{41}{3} = 0$ Solve for $\lambda$: $35\lambda = \frac{41}{3} \Rightarrow \lambda = \frac{41}{3 \times 35} = \frac{41}{105}$

Step 6: Calculate the Coordinates of P and the Final Value Substitute the value of $\lambda = \frac{41}{105}$ back into the parametric coordinates of $P$: $\alpha = -\frac{41}{105} + \frac{4}{3} = \frac{-41 + 4 \times 35}{105} = \frac{-41 + 140}{105} = \frac{99}{105} = \frac{33}{35}$ $\beta = 5\left(\frac{41}{105}\right) - \frac{2}{3} = \frac{41}{21} - \frac{2}{3} = \frac{41 - 2 \times 7}{21} = \frac{41 - 14}{21} = \frac{27}{21} = \frac{9}{7}$ $\gamma = 3\left(\frac{41}{105}\right) = \frac{41}{35}$ So, the foot of the perpendicular is $P\left(\frac{33}{35}, \frac{9}{7}, \frac{41}{35}\right)$.

We need to find the value of $35(\alpha + \beta + \gamma)$. First, calculate the sum $\alpha + \beta + \gamma$: $\alpha + \beta + \gamma = \frac{33}{35} + \frac{9}{7} + \frac{41}{35}$ To add these fractions, find a common denominator, which is 35: $\alpha + \beta + \gamma = \frac{33}{35} + \frac{9 \times 5}{35} + \frac{41}{35} = \frac{33 + 45 + 41}{35} = \frac{119}{35}$ Finally, multiply this sum by 35: $35(\alpha + \beta + \gamma) = 35 \times \frac{119}{35} = 119$

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Common Mistakes & Tips

• Cross Product Errors: Be very careful with signs and calculations when computing the cross product of normal vectors. A single sign error will propagate through the entire solution.
• Arithmetic with Fractions: Calculations involving fractions for the point on the line, $\lambda$, and the final coordinates are prone to errors. Double-check all fraction additions, subtractions, and multiplications.
• Incorrect Direction Vector for Dot Product: Ensure that the direction vector used in the dot product condition $\overrightarrow{AP} \cdot \overrightarrow a = 0$ is indeed the direction vector of the line $L$, not one of the normal vectors of the planes.
• Verification: After finding a point on the line of intersection, quickly substitute its coordinates back into the original plane equations to verify it satisfies both. This catches errors early.

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Summary

This problem required a systematic approach combining two key 3D geometry concepts. We first transformed the plane equations into Cartesian form to easily extract their normal vectors. The cross product of these normal vectors yielded the direction vector of the line of intersection. By setting one coordinate to zero, we found a specific point on this line, allowing us to write its parametric equation. Finally, using the condition that the vector from the external point to the foot of the perpendicular on the line is orthogonal to the line's direction vector, we solved for the parameter $\lambda$. Substituting $\lambda$ back into the parametric equations gave the coordinates of the foot of the perpendicular, and from there, the required value of $35(\alpha + \beta + \gamma)$. Our calculations consistently lead to a value of 119.

The final answer is $\boxed{119}$ which corresponds to option (B).