Key Concepts and Formulas

1. Equation of a Line in 3D: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(a, b, c)$ can be represented in its symmetric form as $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$. Its parametric vector form is $\mathbf{r}(t) = (x_0, y_0, z_0) + t(a, b, c)$, where $t$ is a scalar parameter.
2. Direction Vector and Perpendicularity: The vector $(a, b, c)$ is the direction vector of the line. If two lines are perpendicular, their direction vectors are orthogonal, meaning their dot product is zero.
3. Cross Product: The cross product of two vectors $\mathbf{u}$ and $\mathbf{v}$, denoted $\mathbf{u} \times \mathbf{v}$, yields a vector that is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$. This is fundamental for finding the direction of a line perpendicular to two given lines. For $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$, $\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = (u_2v_3 - u_3v_2)\mathbf{i} - (u_1v_3 - u_3v_1)\mathbf{j} + (u_1v_2 - u_2v_1)\mathbf{k}$

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Step-by-Step Solution

Step 1: Identify Direction Vectors of $L_1$ and $L_2$. We begin by extracting the direction vectors for the given lines $L_1$ and $L_2$ from their symmetric equations. The symmetric form $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ directly provides the direction vector $(a, b, c)$.

For line $\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$: The direction vector for $L_1$ is $\mathbf{u} = (1, -1, 2)$. A point on $L_1$ is $P_1 = (1, 2, 1)$.

For line $\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$: The direction vector for $L_2$ is $\mathbf{v} = (-1, 2, 1)$. A point on $L_2$ is $P_2 = (-1, 2, 0)$.

Reasoning: Identifying these direction vectors is the first crucial step as they define the orientation of the lines in 3D space, which is essential for determining perpendicularity.

Step 2: Determine the Direction Vector of Line $L_3$. The problem states that line $L_3$ is perpendicular to both $L_1$ and $L_2$. A vector that is perpendicular to two other vectors can be found using their cross product. Therefore, the direction vector of $L_3$, let's denote it as $\mathbf{d}$, will be parallel to the cross product of $\mathbf{u}$ and $\mathbf{v}$.

Let's compute $\mathbf{d} = \mathbf{u} \times \mathbf{v}$: $\mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix}$ Expanding the determinant:

• $\mathbf{i}$-component: $(-1)(1) - (2)(2) = -1 - 4 = -5$
• $\mathbf{j}$-component: $-( (1)(1) - (2)(-1) ) = -(1 - (-2)) = -(1+2) = -3$
• $\mathbf{k}$-component: $(1)(2) - (-1)(-1) = 2 - 1 = 1$

So, the direction vector for $L_3$ is $\mathbf{d} = (-5, -3, 1)$. Reasoning: The cross product provides a vector that is inherently orthogonal to both input vectors, perfectly satisfying the perpendicularity condition for $L_3$.

Step 3: Express the Point $(\alpha, \beta, \gamma)$ using the Intersection Condition. We are given that $L_3$ passes through the point $Q(\alpha, \beta, \gamma)$ and intersects $L_1$. Let the point of intersection between $L_3$ and $L_1$ be $A$. Since point $A$ lies on $L_1$, we can represent its coordinates using the parametric form of $L_1$. Using $P_1(1,2,1)$ and $\mathbf{u}=(1,-1,2)$: $A = (1+t, 2-t, 1+2t)$ for some scalar parameter $t$.

Now, line $L_3$ passes through $Q(\alpha, \beta, \gamma)$ and also through $A$. Since $\mathbf{d}=(-5, -3, 1)$ is the direction vector of $L_3$, the vector $\vec{QA}$ (or $\vec{AQ}$) must be parallel to $\mathbf{d}$. This means that $Q$ can be expressed as $A$ plus some scalar multiple of $\mathbf{d}$. Let this scalar multiple be $s$: $Q = A + s\mathbf{d}$ Substituting the coordinates of $A$ and the vector $\mathbf{d}$: $(\alpha, \beta, \gamma) = (1+t, 2-t, 1+2t) + s(-5, -3, 1)$ This gives us a system of equations for $\alpha, \beta, \gamma$ in terms of parameters $t$ and $s$: $\begin{cases} \alpha = 1 + t - 5s \\ \beta = 2 - t - 3s \\ \gamma = 1 + 2t + s \end{cases}$ Reasoning: This step leverages the fact that $L_3$ intersects $L_1$. By parameterizing the intersection point $A$ on $L_1$ and then using the direction vector of $L_3$, we can express the coordinates of $Q(\alpha, \beta, \gamma)$ in terms of two parameters $t$ and $s$. This formulation is key to evaluating the target expression.

Step 4: Evaluate the Expression $|5\alpha - 11\beta - 8\gamma|$. Now, we substitute the expressions for $\alpha, \beta, \gamma$ from Step 3 into the target expression: $5\alpha - 11\beta - 8\gamma = 5(1+t-5s) - 11(2-t-3s) - 8(1+2t+s)$ Let's carefully expand and simplify the expression: $= (5 + 5t - 25s) - (22 - 11t - 33s) - (8 + 16t + 8s)$ Distribute the negative signs: $= 5 + 5t - 25s - 22 + 11t + 33s - 8 - 16t - 8s$ Now, group the constant terms, terms involving $t$, and terms involving $s$:

• Constant terms: $5 - 22 - 8 = -17 - 8 = -25$
• Terms with $t$: $5t + 11t - 16t = 16t - 16t = 0t = 0$
• Terms with $s$: $-25s + 33s - 8s = 8s - 8s = 0s = 0$

Combining these, we find: $5\alpha - 11\beta - 8\gamma = -25 + 0 + 0 = -25$ Finally, we need to find the absolute value of this expression: $|5\alpha - 11\beta - 8\gamma| = |-25| = 25$ Reasoning: This is the final computational step. The cancellation of parameters $t$ and $s$ indicates that the value of the expression is independent of the specific point of intersection or the exact location of $Q$ along $L_3$, as long as the conditions of the problem are met. This is a common pattern in JEE problems designed to simplify calculations.

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Common Mistakes & Tips

• Cross Product Sign Errors: Be extremely cautious with the signs when calculating the cross product, especially for the $\mathbf{j}$ component (which has a negative sign in the determinant expansion).
• Interpreting Line Equations: Ensure correct extraction of points and direction vectors from the symmetric form. For example, $x+1$ corresponds to $x-(-1)$, so $x_0 = -1$.
• Algebraic Precision: The final substitution and simplification step requires meticulous attention to signs and distribution to avoid arithmetic errors.
• Independence of Parameters: If parameters like $t$ and $s$ cancel out in the final expression, it's often a good sign that your calculations are correct, as it implies the quantity being sought is invariant under the specific choices of these parameters.

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Summary

This problem required a systematic application of 3D geometry concepts. We first identified the direction vectors of the given lines $L_1$ and $L_2$. Then, we used the cross product of these vectors to determine the direction vector of $L_3$, which is perpendicular to both. By leveraging the condition that $L_3$ passes through $(\alpha, \beta, \gamma)$ and intersects $L_1$, we expressed $\alpha, \beta, \gamma$ in terms of parameters $t$ and $s$. Substituting these expressions into the target algebraic expression, we found that all parametric terms canceled out, yielding a constant value. The absolute value of this constant gave the final answer.

The final answer is $\boxed{25}$ which corresponds to option (A).