1. Key Concepts and Formulas

• Symmetric Form of a Straight Line: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
• Parametric Form of a Straight Line: By equating the symmetric form to a parameter (e.g., $\lambda$), any general point on the line can be represented as $(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$. This form is essential for finding intersection points.
• Point of Intersection of Two Lines: If two lines intersect, a unique point exists that lies on both. We find this by setting the general points (using different parameters) from each line equal to each other and solving the resulting system of equations.
• Condition for a Point to Lie on a Plane: A point $(x_0, y_0, z_0)$ lies on a plane $Ax + By + Cz = D$ if its coordinates satisfy the plane's equation: $Ax_0 + By_0 + Cz_0 = D$.

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2. Step-by-Step Solution

The problem asks for the value of $\alpha - \beta$, given two lines whose intersection point lies on a specific plane. Our strategy involves finding the general points on each line, determining their intersection point, and then using the plane condition to solve for $\alpha$ and $\beta$.

Step 1: Representing Lines in Parametric Form To find the point of intersection, we express any general point on each line using a parameter.

Line 1 ($L_1$): $\frac{x - \alpha}{1} = \frac{y - 1}{2} = \frac{z - 1}{3}$ Let's equate this to a parameter, say $\phi$: $\frac{x - \alpha}{1} = \frac{y - 1}{2} = \frac{z - 1}{3} = \phi$ From this, the coordinates of any point $P_1$ on Line 1 are: $x = \phi + \alpha$ $y = 2\phi + 1$ $z = 3\phi + 1$ So, a general point on Line 1 is $P_1(\phi + \alpha, 2\phi + 1, 3\phi + 1)$.

Line 2 ($L_2$): $\frac{x - 4}{\beta} = \frac{y - 6}{3} = \frac{z - 7}{3}$ Let's equate this to a different parameter, say $q$: $\frac{x - 4}{\beta} = \frac{y - 6}{3} = \frac{z - 7}{3} = q$ From this, the coordinates of any point $P_2$ on Line 2 are: $x = q\beta + 4$ $y = 3q + 6$ $z = 3q + 7$ So, a general point on Line 2 is $P_2(q\beta + 4, 3q + 6, 3q + 7)$.

Step 2: Finding the Point of Intersection For the two lines to intersect, there must be a common point. This means that for some specific values of $\phi$ and $q$, the coordinates of $P_1$ and $P_2$ must be identical. We equate the corresponding coordinates:

1. x-coordinate: $\phi + \alpha = q\beta + 4 \quad \ldots(1)$
2. y-coordinate: $2\phi + 1 = 3q + 6 \quad \ldots(2)$
3. z-coordinate: $3\phi + 1 = 3q + 7 \quad \ldots(3)$

We first solve equations (2) and (3) for $\phi$ and $q$, as they only involve these two parameters: From equation (2): $2\phi - 3q = 5$ From equation (3): $3\phi - 3q = 6$

Subtract equation (2) from equation (3): $(3\phi - 3q) - (2\phi - 3q) = 6 - 5$ $\phi = 1$ Now substitute $\phi = 1$ into equation (2): $2(1) - 3q = 5$ $2 - 3q = 5$ $-3q = 3$ $q = -1$

So, the lines intersect when $\phi = 1$ and $q = -1$. Now, we find the coordinates of the intersection point, let's call it $I$. Using $P_1$ with $\phi=1$: $x_I = 1 + \alpha$ $y_I = 2(1) + 1 = 3$ $z_I = 3(1) + 1 = 4$ So, the point of intersection is $I(1 + \alpha, 3, 4)$.

We can also use $P_2$ with $q=-1$ to verify and establish a relationship between $\alpha$ and $\beta$: $x_I = (-1)\beta + 4 = 4 - \beta$ $y_I = 3(-1) + 6 = 3$ $z_I = 3(-1) + 7 = 4$ Equating the $x$-coordinates from both expressions for $I$: $1 + \alpha = 4 - \beta$ $\alpha + \beta = 3 \quad \ldots(4)$ This is a crucial relationship between $\alpha$ and $\beta$.

Step 3: Applying the Plane Condition The problem states that the point of intersection $I(1 + \alpha, 3, 4)$ lies on the plane $x + 2y - z = 8$. This means the coordinates of $I$ must satisfy the plane's equation.

Substitute $x = 1 + \alpha$, $y = 3$, and $z = 4$ into the plane equation: $(1 + \alpha) + 2(3) - (4) = 8$ $1 + \alpha + 6 - 4 = 8$ $\alpha + 3 = 8$ $\alpha = 8 - 3$ $\alpha = 5$ We have found the value of $\alpha$.

Step 4: Calculating $\alpha - \beta$ Now that we know $\alpha = 5$, we can use the relationship from equation (4), $\alpha + \beta = 3$, to find $\beta$. $5 + \beta = 3$ $\beta = 3 - 5$ $\beta = -2$ This value $\beta = -2$ satisfies the condition $\beta \ne 0$.

Finally, we need to calculate $\alpha - \beta$: $\alpha - \beta = 5 - (-2)$ $\alpha - \beta = 5 + 2$ $\alpha - \beta = 7$

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3. Common Mistakes & Tips

• Using the Same Parameter: Always use different parameters (e.g., $\phi$ and $q$) for different lines when finding an intersection point. Using the same parameter implies the lines are identical or the parameter value is fixed, which is generally incorrect.
• Algebraic Errors: Be careful with signs and basic arithmetic when solving systems of equations and substituting values. A small error can propagate through the entire solution.
• Verifying Coordinates: After finding the parameter values for intersection, substitute them back into both parametric forms to ensure all coordinates match. This helps catch errors early.

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4. Summary

We began by converting the given symmetric equations of the lines into their parametric forms to represent general points on each line. By equating the corresponding coordinates of these general points, we formed a system of equations. Solving this system yielded the specific parameter values ($\phi=1, q=-1$) that define the intersection point, leading to the coordinates $I(1+\alpha, 3, 4)$ and the relationship $\alpha + \beta = 3$. Finally, by substituting the coordinates of the intersection point into the equation of the given plane, we determined $\alpha = 5$. Using this value, we found $\beta = -2$. The required value $\alpha - \beta$ was then calculated as $7$.

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5. Final Answer

The final answer is $\boxed{7}$, which corresponds to option (D).