Key Concepts and Formulas

1. Equation of a Plane: A plane can be uniquely determined by a point on it and a vector normal to it. If a plane passes through a line $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$ and an external point $P_2(x_2, y_2, z_2)$:

   • The line provides a point $P_1(x_1, y_1, z_1)$ and a direction vector $\vec{d} = l\hat{i} + m\hat{j} + n\hat{k}$, both lying in the plane.
   • The vector $\vec{P_1P_2} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$ also lies in the plane.
   • The normal vector to the plane, $\vec{n}$, is perpendicular to both $\vec{d}$ and $\vec{P_1P_2}$. Hence, $\vec{n} = \vec{d} \times \vec{P_1P_2}$.
   • The equation of the plane in point-normal form is $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$, where $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ and $(x_0, y_0, z_0)$ is any point on the plane (e.g., $P_1$ or $P_2$).

2. Image of a Point in a Plane: The image $P'(x', y', z')$ of a point $P_0(x_0, y_0, z_0)$ in the plane $Ax+By+Cz+D=0$ satisfies two conditions: the line segment $P_0P'$ is perpendicular to the plane, and its midpoint lies on the plane. The coordinates of the image point are given by the formula: $\frac{x' - x_0}{A} = \frac{y' - y_0}{B} = \frac{z' - z_0}{C} = \frac{-2(Ax_0+By_0+Cz_0+D)}{A^2+B^2+C^2}$

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Step-by-Step Solution

Step 1: Determine the Equation of Plane P

Our first objective is to find the algebraic equation of plane P using the given line and point.

• 1.1 Extracting Information from the Line and Given Point: The line is given by $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$.

  • Why: To define the plane, we need specific geometric elements. The symmetric form of the line directly provides a point on the line and its direction. From the line, we identify:
  • A point on the line, $P_1$: $(1, 2, -5)$.
  • The direction vector of the line, $\vec{d}$: $\langle 1, -3, 7 \rangle$. The plane P also passes through an additional point, $P_2$: $(2, 4, -3)$.

• 1.2 Forming a Vector Lying in the Plane: Since both $P_1(1,2,-5)$ and $P_2(2,4,-3)$ lie in the plane, the vector connecting them must also lie within the plane.

  • Why: To find the normal vector using the cross product, we require two distinct, non-parallel vectors that lie within the plane. We already have $\vec{d}$, and this step gives us a second such vector. Let's calculate the vector $\vec{P_1P_2}$: $\vec{P_1P_2} = (2-1)\hat{i} + (4-2)\hat{j} + (-3-(-5))\hat{k}$ $\vec{P_1P_2} = 1\hat{i} + 2\hat{j} + 2\hat{k}$

• 1.3 Calculating the Normal Vector to the Plane: The normal vector $\vec{n}$ to the plane is perpendicular to any two non-parallel vectors lying in the plane. We have two such vectors: $\vec{d} = \langle 1, -3, 7 \rangle$ and $\vec{P_1P_2} = \langle 1, 2, 2 \rangle$. Therefore, we find $\vec{n}$ by taking their cross product.

  • Why: The cross product intrinsically produces a vector that is orthogonal to both input vectors. This resulting vector serves as the normal vector for the plane containing the two original vectors. $\vec{n} = \vec{d} \times \vec{P_1P_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 7 \\ 1 & 2 & 2 \end{vmatrix}$ Expanding the determinant: $\vec{n} = ((-3)(2) - (7)(2))\hat{i} - ((1)(2) - (7)(1))\hat{j} + ((1)(2) - (-3)(1))\hat{k}$ $\vec{n} = (-6 - 14)\hat{i} - (2 - 7)\hat{j} + (2 + 3)\hat{k}$ $\vec{n} = -20\hat{i} + 5\hat{j} + 5\hat{k}$ For simplicity in subsequent calculations, we can divide this normal vector by a common factor of $5$: $\vec{n'} = -4\hat{i} + \hat{j} + \hat{k}$ It is also common practice to choose a normal vector with a positive leading coefficient (though not strictly necessary). Multiplying by $-1$: $\vec{n''} = 4\hat{i} - \hat{j} - \hat{k}$ Thus, the coefficients for our plane equation will be $A=4, B=-1, C=-1$.

• 1.4 Writing the Equation of Plane P: With the normal vector $\vec{n''} = \langle 4, -1, -1 \rangle$ and a point on the plane (let's use $P_1(1,2,-5)$), we can use the point-normal form of the plane equation.

  • Why: This is the standard and most direct way to construct the equation of a plane once its normal vector and at least one point on it are known. The equation is $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$: $4(x-1) + (-1)(y-2) + (-1)(z-(-5)) = 0$ $4(x-1) - (y-2) - (z+5) = 0$ $4x - 4 - y + 2 - z - 5 = 0$ $4x - y - z - 7 = 0$ This is the equation of plane P.

Step 2: Find the Image of the Point $(-1,3,4)$ in Plane P

Now, we will determine the image of the given point in the plane $P: 4x-y-z-7=0$.

• 2.1 Identifying Parameters: Let the given point be $Q(x_0, y_0, z_0) = (-1,3,4)$, and its image be $Q'(\alpha, \beta, \gamma)$. The equation of plane P is $4x-y-z-7=0$.

  • Why: Clearly identifying each parameter from the point and plane equation is critical for accurate substitution into the image formula. Comparing with the general plane form $Ax+By+Cz+D=0$:
  • $x_0 = -1, y_0 = 3, z_0 = 4$
  • $A = 4, B = -1, C = -1, D = -7$

• 2.2 Applying the Image Formula: We use the formula for the coordinates of the image point: $\frac{\alpha - x_0}{A} = \frac{\beta - y_0}{B} = \frac{\gamma - z_0}{C} = \frac{-2(Ax_0+By_0+Cz_0+D)}{A^2+B^2+C^2}$

  • Why: This formula directly provides the coordinates of the image point without needing to derive the line perpendicular to the plane or find the midpoint. First, calculate the term $Ax_0+By_0+Cz_0+D$: $4(-1) + (-1)(3) + (-1)(4) + (-7)$ $= -4 - 3 - 4 - 7 = -18$ Next, calculate the term $A^2+B^2+C^2$: $(4)^2 + (-1)^2 + (-1)^2$ $= 16 + 1 + 1 = 18$ Now substitute these values into the image formula: $\frac{\alpha - (-1)}{4} = \frac{\beta - 3}{-1} = \frac{\gamma - 4}{-1} = \frac{-2(-18)}{18}$ $\frac{\alpha + 1}{4} = \frac{\beta - 3}{-1} = \frac{\gamma - 4}{-1} = \frac{36}{18}$ $\frac{\alpha + 1}{4} = \frac{\beta - 3}{-1} = \frac{\gamma - 4}{-1} = 2$

• 2.3 Solving for $\alpha, \beta, \gamma$: Equating each part of the fraction to $2$ to find the coordinates of the image:

  • For $\alpha$: $\frac{\alpha + 1}{4} = 2 \implies \alpha + 1 = 8 \implies \alpha = 7$
  • For $\beta$: $\frac{\beta - 3}{-1} = 2 \implies \beta - 3 = -2 \implies \beta = 1$
  • For $\gamma$: $\frac{\gamma - 4}{-1} = 2 \implies \gamma - 4 = -2 \implies \gamma = 2$ So, the image of the point $(-1,3,4)$ in plane P is $(\alpha, \beta, \gamma) = (7, 1, 2)$.

Step 3: Calculate $\alpha+\beta+\gamma$

The problem asks for the sum of the coordinates of the image point.

• 3.1 Summing the Coordinates:
  • Why: This is the final step as per the problem statement's requirement. We found $\alpha=7$, $\beta=1$, and $\gamma=2$. $\alpha + \beta + \gamma = 7 + 1 + 2 = 10$

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Common Mistakes & Tips

• Sign Errors: Be extremely cautious with negative signs, especially during cross product calculations, vector subtractions, and substitutions into the image formula. A single sign error can propagate and lead to an incorrect final answer.
• Arithmetic Precision: Double-check all arithmetic operations. Simple calculation mistakes are common under exam pressure.
• Simplifying Normal Vector: Always simplify the normal vector by dividing by common factors. This reduces the magnitude of the numbers involved, making subsequent calculations (like those in the image formula) less prone to errors.
• Choice of Point for Plane Equation: Remember that any point known to lie on the plane (e.g., $P_1$ or $P_2$) can be used in the point-normal form $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$. Both will yield the same final plane equation.

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Summary

This problem required a two-stage approach: first, determining the equation of the plane, and then finding the image of a point in that plane. We started by extracting a point and direction vector from the given line, then formed a second vector using the line's point and the external point. The cross product of these two vectors yielded the normal vector to the plane. Using this normal vector and one of the points, we constructed the plane's equation. Finally, we applied the standard formula for the image of a point in a plane to find the coordinates $(\alpha, \beta, \gamma)$ and summed them to get the required value.

The final answer is $\boxed{10}$, which corresponds to option (A).