Key Concepts and Formulas

1. Perpendicular Lines in 3D Geometry: Two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are perpendicular if and only if the dot product of their corresponding direction ratios is zero: $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
2. Equation of a Line in Symmetric and Parametric Form: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(L, M, N)$ can be written in symmetric form as $\frac{x-x_1}{L} = \frac{y-y_1}{M} = \frac{z-z_1}{N}$. By equating this to a parameter $\lambda$, we obtain the parametric form of any point on the line: $(x_1 + L\lambda, y_1 + M\lambda, z_1 + N\lambda)$.
3. Intersection of a Line and a Plane: To find the point where a line intersects a plane, substitute the parametric coordinates of a general point on the line into the equation of the plane. Solving for the parameter $\lambda$ will yield the specific value corresponding to the intersection point.

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Step-by-Step Solution

Step 1: Determine the values of $a$ and $b$ using the perpendicularity condition.

We are given a line $L_1$ with direction ratios $(a, -4a, -7)$. This line $L_1$ is perpendicular to two other lines:

• $L_2$ with direction ratios $(3, -1, 2b)$
• $L_3$ with direction ratios $(b, a, -2)$

Reasoning: Since two lines are perpendicular, the dot product of their direction ratios must be zero. We apply this condition to $L_1$ and $L_2$, and then to $L_1$ and $L_3$.

For $L_1 \perp L_2$: $(a)(3) + (-4a)(-1) + (-7)(2b) = 0$ $3a + 4a - 14b = 0$ $7a - 14b = 0$ Dividing by 7, we get: $a - 2b = 0 \implies a = 2b \quad \text{(Equation 1)}$

For $L_1 \perp L_3$: $(a)(b) + (-4a)(a) + (-7)(-2) = 0$ $ab - 4a^2 + 14 = 0 \quad \text{(Equation 2)}$

Now, we solve the system of equations (1) and (2) for $a$ and $b$. Substitute $a = 2b$ from Equation 1 into Equation 2: $(2b)b - 4(2b)^2 + 14 = 0$ $2b^2 - 4(4b^2) + 14 = 0$ $2b^2 - 16b^2 + 14 = 0$ $-14b^2 + 14 = 0$ $14b^2 = 14$ $b^2 = 1$ From $b^2 = 1$, we have $b = 1$ or $b = -1$.

If $b=1$, then from Equation 1, $a = 2(1) = 2$. If $b=-1$, then from Equation 1, $a = 2(-1) = -2$.

In both cases, we find $a^2 = (\pm 2)^2 = 4$ and $b^2 = (\pm 1)^2 = 1$. These values are consistent and uniquely determine $a^2$ and $b^2$.

Step 2: Formulate the equation of the given line in symmetric and parametric form.

The line is given by $\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$.

Reasoning: We substitute the determined values of $a^2$ and $b^2$ into the line equation to get its specific form. Then, we introduce a parameter to express any point on the line.

First, calculate the denominators: $a^2 + b^2 = 4 + 1 = 5$ $a^2 - b^2 = 4 - 1 = 3$

Substitute these values into the line equation: $\frac{x+1}{5} = \frac{y-2}{3} = \frac{z}{1}$ This is the symmetric form of the line.

To express a general point on this line, we equate the ratios to a parameter $\lambda$: $\frac{x+1}{5} = \frac{y-2}{3} = \frac{z}{1} = \lambda$ From this, we can write $x, y, z$ in terms of $\lambda$: $x+1 = 5\lambda \implies x = 5\lambda - 1$ $y-2 = 3\lambda \implies y = 3\lambda + 2$ $z = 1\lambda \implies z = \lambda$ So, any point on the line can be represented as $(5\lambda - 1, 3\lambda + 2, \lambda)$. This is the parametric form of the line.

Step 3: Find the point of intersection of the line and the plane.

The equation of the plane is $x-y+z=0$. The general point on the line is $(5\lambda - 1, 3\lambda + 2, \lambda)$.

Reasoning: The point of intersection must satisfy both the line's equation (already incorporated into the general point) and the plane's equation. We substitute the parametric coordinates of the general point into the plane equation to find the value of $\lambda$ at the intersection.

Substitute the parametric form of the line into the plane equation: $(5\lambda - 1) - (3\lambda + 2) + (\lambda) = 0$ Now, solve this linear equation for $\lambda$: $5\lambda - 1 - 3\lambda - 2 + \lambda = 0$ $(5\lambda - 3\lambda + \lambda) + (-1 - 2) = 0$ $3\lambda - 3 = 0$ $3\lambda = 3$ $\lambda = 1$

Now, substitute $\lambda=1$ back into the parametric form of the line to find the coordinates of the intersection point $(\alpha, \beta, \gamma)$: $\alpha = 5(1) - 1 = 4$ $\beta = 3(1) + 2 = 5$ $\gamma = 1$ Thus, the point of intersection $(\alpha, \beta, \gamma)$ is $(4, 5, 1)$.

Step 4: Calculate the required sum $\alpha+\beta+\gamma$.

We found the point of intersection to be $(\alpha, \beta, \gamma) = (4, 5, 1)$.

Reasoning: The problem asks for the sum of these coordinates. $\alpha + \beta + \gamma = 4 + 5 + 1$ $\alpha + \beta + \gamma = 10$

(Self-correction: The problem's "Correct Answer" is given as 3. However, a meticulous step-by-step derivation based on the provided question text consistently leads to 10. As an expert, I must present the mathematically correct derivation for the given problem statement. It is possible there is a discrepancy between the problem statement and the intended answer in the source material.)

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Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with signs and arithmetic, especially when solving systems of equations and substituting values. A single sign error can propagate and lead to an incorrect final answer.
• Misinterpreting Direction Ratios: Remember that direction ratios are proportional to direction cosines. The dot product condition for perpendicularity ($l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$) is fundamental and must be applied correctly.
• Incorrect Parametric Form: Ensure that when converting the symmetric form of a line to its parametric form, the signs and constants are correctly handled (e.g., $x+1$ becomes $x = L\lambda - 1$, not $L\lambda + 1$).
• Verification: After finding the intersection point, it's a good practice to substitute its coordinates back into both the line equation and the plane equation to ensure it satisfies both.

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Summary

This problem effectively tests multiple core concepts in 3D geometry. We began by using the perpendicularity condition for lines to establish a system of equations for the unknown parameters $a$ and $b$. Solving this system yielded unique values for $a^2$ and $b^2$. These values were then used to define the equation of the given line in its symmetric and parametric forms. Finally, by substituting the parametric coordinates of a general point on the line into the plane equation, we determined the point of intersection. The sum of the coordinates of this intersection point was then calculated.

The final answer is $\boxed{10}$.