This problem requires us to find the foot of a perpendicular from a given point to a line, where the line itself is the intersection of two planes. We will systematically approach this using concepts from 3D geometry.

1. Key Concepts and Formulas

• Equation of a Plane: A plane can be defined by a point it passes through and its normal vector. If the plane passes through points $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$, its equation can be found using the determinant form or by finding two vectors in the plane, taking their cross product to get the normal vector, and then using the point-normal form.
• Equation of a Line of Intersection: The direction vector of the line of intersection of two planes is perpendicular to both plane normal vectors. Thus, it can be found by taking the cross product of the normal vectors of the two planes. A point on the line can be found by setting one variable to zero (or any constant) in both plane equations and solving for the other two.
• Foot of the Perpendicular from a Point to a Line: If $A$ is the given point and $L$ is the line with direction vector $\vec{v}$, any point $F$ on $L$ can be expressed parametrically. The vector $\vec{AF}$ (from point $A$ to point $F$) must be perpendicular to the direction vector $\vec{v}$ of the line. Therefore, their dot product $\vec{AF} \cdot \vec{v}$ must be zero. This condition allows us to find the parameter value for $F$, and subsequently its coordinates.

2. Step-by-Step Solution

Step 1: Find the equation of Plane $P_2$.

• What we are doing: We need the algebraic equation of plane $P_2$ to later find its intersection with $P_1$.
• Why: The line of intersection requires the equations of both planes.
• Math: Plane $P_2$ passes through points $B(2,-1,0)$, $C(2,0,-1)$, and $D(5,1,1)$. First, we find two vectors lying in the plane: $\vec{BC} = C - B = (2-2, 0-(-1), -1-0) = (0, 1, -1)$ $\vec{BD} = D - B = (5-2, 1-(-1), 1-0) = (3, 2, 1)$ The normal vector $\vec{n_2}$ to plane $P_2$ is the cross product of $\vec{BC}$ and $\vec{BD}$: $\vec{n_2} = \vec{BC} \times \vec{BD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & -1 \\ 3 & 2 & 1 \end{vmatrix}$ $\vec{n_2} = \mathbf{i}((1)(1) - (-1)(2)) - \mathbf{j}((0)(1) - (-1)(3)) + \mathbf{k}((0)(2) - (1)(3))$ $\vec{n_2} = \mathbf{i}(1+2) - \mathbf{j}(0+3) + \mathbf{k}(0-3) = (3, -3, -3)$ We can use a simpler normal vector parallel to $(3,-3,-3)$, which is $(1,-1,-1)$. So, we use $\vec{n_2} = (1, -1, -1)$. Using the point-normal form with point $B(2,-1,0)$ and $\vec{n_2}=(1,-1,-1)$: $1(x-2) - 1(y-(-1)) - 1(z-0) = 0$ $x-2 - (y+1) - z = 0$ $x - y - z - 3 = 0$ Thus, the equation of plane $P_2$ is $x - y - z = 3$.

Step 2: Find the equation of the Line of Intersection ($L$) of $P_1$ and $P_2$.

• What we are doing: We are determining the parametric equation of the line where the two planes $P_1$ and $P_2$ intersect.

• Why: The foot of the perpendicular will lie on this line, so its equation is essential.

• Math: The equations of the two planes are: $P_1: 3x - y - 7z = 11$ (Normal vector $\vec{n_1} = (3, -1, -7)$) $P_2: x - y - z = 3$ (Normal vector $\vec{n_2} = (1, -1, -1)$) The direction vector $\vec{v}$ of the line of intersection is perpendicular to both $\vec{n_1}$ and $\vec{n_2}$. We find it using their cross product: $\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & -7 \\ 1 & -1 & -1 \end{vmatrix}$ $\vec{v} = \mathbf{i}((-1)(-1) - (-7)(-1)) - \mathbf{j}((3)(-1) - (-7)(1)) + \mathbf{k}((3)(-1) - (-1)(1))$ $\vec{v} = \mathbf{i}(1-7) - \mathbf{j}(-3+7) + \mathbf{k}(-3+1)$ $\vec{v} = (-6, -4, -2)$ We can use a simpler direction vector parallel to $(-6,-4,-2)$, which is $(3,2,1)$. So, let $\vec{v} = (3,2,1)$.

  Next, we need a point on the line of intersection. We can find this by setting one coordinate to zero (e.g., $z=0$) in both plane equations and solving for $x$ and $y$: For $z=0$: $3x - y = 11 \quad (1)$ $x - y = 3 \quad (2)$ Subtracting equation (2) from equation (1): $(3x - y) - (x - y) = 11 - 3$ $2x = 8 \implies x = 4$ Substitute $x=4$ into equation (2): $4 - y = 3 \implies y = 1$ So, a point on the line of intersection is $P_0(4,1,0)$. The parametric equation of the line of intersection $L$ is: $x = 4 + 3t$ $y = 1 + 2t$ $z = t$ or $\vec{r}(t) = (4+3t, 1+2t, t)$.

Step 3: Find the foot of the perpendicular $F(\alpha, \beta, \gamma)$ from point $A(7,4,-1)$ to line $L$.

• What we are doing: We are identifying the specific point on line $L$ that is closest to point $A$.
• Why: This point is defined as the foot of the perpendicular, and its coordinates are needed to calculate $\alpha+\beta+\gamma$.
• Math: Let $F(\alpha, \beta, \gamma)$ be the foot of the perpendicular from $A(7,4,-1)$ to the line $L$. Any point on line $L$ can be represented as $F(4+3t, 1+2t, t)$ for some scalar $t$. The vector $\vec{AF}$ connects point $A(7,4,-1)$ to point $F(4+3t, 1+2t, t)$: $\vec{AF} = (4+3t - 7, 1+2t - 4, t - (-1))$ $\vec{AF} = (3t-3, 2t-3, t+1)$ The direction vector of the line $L$ is $\vec{v} = (3,2,1)$. Since $\vec{AF}$ must be perpendicular to $\vec{v}$, their dot product must be zero: $\vec{AF} \cdot \vec{v} = 0$ $(3t-3)(3) + (2t-3)(2) + (t+1)(1) = 0$ $9t - 9 + 4t - 6 + t + 1 = 0$ $14t - 14 = 0$ $14t = 14$ $t = 1$ Now substitute $t=1$ back into the coordinates of $F$: $\alpha = 4 + 3(1) = 7$ $\beta = 1 + 2(1) = 3$ $\gamma = 1$ So, the foot of the perpendicular is $F(7,3,1)$.

Step 4: Calculate $\alpha+\beta+\gamma$.

• What we are doing: Summing the coordinates of the foot of the perpendicular.
• Why: This is the final value requested by the problem.
• Math: $\alpha+\beta+\gamma = 7+3+1 = 11$

Common Mistakes & Tips

• Sign Errors in Cross Products: Be very careful with signs when calculating cross products for normal or direction vectors. A single sign error can propagate through the entire problem.
• Arithmetic Errors: Double-check additions, subtractions, and multiplications, especially when dealing with negative numbers. A small arithmetic mistake can lead to an incorrect value for $t$.
• Choosing a Point on the Line: When finding a point on the line of intersection, try setting $z=0$ (or $x=0$ or $y=0$) and solve for the other two variables. If this results in a system with no solution or an undefined point, choose a different variable to set to zero.
• Perpendicularity Condition: Remember that the vector from the external point to the foot of the perpendicular on the line must be orthogonal to the direction vector of the line. This is the key to finding the parameter $t$.

Summary

We first determined the equation of plane $P_2$ using the three given points and found its normal vector. Then, we found the direction vector of the line of intersection of $P_1$ and $P_2$ by taking the cross product of their normal vectors, and identified a point on this line. This allowed us to write the parametric equation of the line of intersection. Finally, we used the condition that the vector from the given point $A$ to the foot of the perpendicular $F$ is orthogonal to the line's direction vector to solve for the parameter $t$. Substituting $t$ back into the line's equation gave us the coordinates of $F(\alpha, \beta, \gamma)$. The sum $\alpha+\beta+\gamma$ was calculated as $7+3+1=11$.

The final answer is $\boxed{11}$.