1. Key Concepts and Formulas

To solve this problem, we'll utilize the following fundamental concepts regarding lines in three-dimensional space:

• Standard (Symmetric) Form of a Line: A line passing through a point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$ is represented as: ${{x - x_1} \over a} = {{y - y_1} \over b} = {{z - z_1} \over c}$ The vector $\vec{d} = (a, b, c)$ is known as the direction vector of the line. It indicates the direction in which the line extends. It is crucial that the coefficients of $x, y, z$ in the numerator are $1$.

• Condition for Perpendicular Lines: Two lines are perpendicular if and only if their direction vectors are orthogonal. If the direction vectors are $\vec{d_1} = (a_1, b_1, c_1)$ and $\vec{d_2} = (a_2, b_2, c_2)$, then their dot product must be zero: $\vec{d_1} \cdot \vec{d_2} = a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

• Angle Between Two Lines: The acute angle $\theta$ between two lines with direction vectors $\vec{d_1} = (a_1, b_1, c_1)$ and $\vec{d_2} = (a_2, b_2, c_2)$ is given by the formula: $\cos \theta = {{|\vec{d_1} \cdot \vec{d_2}|} \over {||\vec{d_1}|| \cdot ||\vec{d_2}||}} = {{|a_1 a_2 + b_1 b_2 + c_1 c_2|} \over {\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}}$ We use the absolute value in the numerator to ensure we find the acute angle between the lines (i.e., $\theta \in [0, \pi/2]$).

2. Step-by-Step Solution

Step 1: Standardize Line Equations and Identify Direction Vectors

The first crucial step is to rewrite all line equations in their standard symmetric form to correctly identify their direction vectors. This is essential because the direction ratios directly correspond to the denominators in the standard form after ensuring the numerators are $(x-x_1)$, $(y-y_1)$, and $(z-z_1)$.

• For line $l_1$: ${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$

  • Reasoning: The condition $z=2$ implies that the $z$-coordinate is constant along the line. This means the line lies in the plane $z=2$, and any change in $z$ along the line is zero. In the standard form, this translates to a $z$-direction ratio of $0$, which can be written as {{z - 2} \over 0}}.
  • Rewriting $l_1$ in standard form: ${{x - 2} \over 3} = {{y - (-1)} \over {-2}} = {{z - 2} \over 0}$
  • The direction vector for $l_1$ is $\vec{d_1} = (3, -2, 0)$.

• For line $l_2$: ${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$

  • Reasoning: The term {{2y + 3} \over \alpha }} is not in the standard form {{y - y_1} \over b}} because the coefficient of $y$ in the numerator is $2$, not $1$. We must factor out the coefficient of $y$ from the numerator and adjust the denominator accordingly.
  • Rewrite the $y$-term: ${{2y + 3} \over \alpha } = {{2(y + {3 \over 2})} \over \alpha } = {{y + {3 \over 2}} \over {{\alpha \over 2}}}$
  • Rewriting $l_2$ in standard form: ${{x - 1} \over 1} = {{y - (-{3 \over 2})} \over {{\alpha \over 2}}} = {{z - (-5)} \over 2}$
  • The direction vector for $l_2$ is $\vec{d_2} = (1, {\alpha \over 2}, 2)$.

• For line $l_3$: ${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$

  • Reasoning: Similar to $l_2$, the terms {{1 - x} \over 3}} and ${{2y - 1} \over { - 4}}$ are not in standard form. For the $x$-term, we need $x - x_1$, so we factor out $-1$. For the $y$-term, we factor out $2$.
  • Rewrite the $x$-term: ${{1 - x} \over 3} = {{-(x - 1)} \over 3} = {{x - 1} \over {-3}}$
  • Rewrite the $y$-term: ${{2y - 1} \over { - 4}} = {{2(y - {1 \over 2})} \over { - 4}} = {{y - {1 \over 2}} \over { - {4 \over 2}}} = {{y - {1 \over 2}} \over {-2}}$
  • The $z$-term $z/4$ is already in standard form, $(z-0)/4$.
  • Rewriting $l_3$ in standard form: ${{x - 1} \over {-3}} = {{y - {1 \over 2}} \over {-2}} = {{z - 0} \over 4}$
  • The direction vector for $l_3$ is $\vec{d_3} = (-3, -2, 4)$.

Step 2: Use the Perpendicularity Condition of $l_1$ and $l_2$ to Find $\alpha$

We are given that lines $l_1$ and $l_2$ are perpendicular. We will use the condition $\vec{d_1} \cdot \vec{d_2} = 0$ to find the value of $\alpha$.

• Direction vector of $l_1$: $\vec{d_1} = (3, -2, 0)$

• Direction vector of $l_2$: $\vec{d_2} = (1, {\alpha \over 2}, 2)$

• Apply the perpendicularity condition: $\vec{d_1} \cdot \vec{d_2} = (3)(1) + (-2)\left({\alpha \over 2}\right) + (0)(2) = 0$ $3 - \alpha + 0 = 0$ $3 - \alpha = 0$ $\alpha = 3$

• Now we have the complete direction vector for $l_2$: $\vec{d_2} = (1, {3 \over 2}, 2)$.

Step 3: Calculate the Angle Between Lines $l_2$ and $l_3$

Now that we have the value of $\alpha$, we can find the angle between $l_2$ and $l_3$ using their direction vectors $\vec{d_2}$ and $\vec{d_3}$.

• Direction vector of $l_2$: $\vec{d_2} = (1, {3 \over 2}, 2)$

• Direction vector of $l_3$: $\vec{d_3} = (-3, -2, 4)$

• Calculate the dot product $\vec{d_2} \cdot \vec{d_3}$: $\vec{d_2} \cdot \vec{d_3} = (1)(-3) + \left({3 \over 2}\right)(-2) + (2)(4)$ $= -3 - 3 + 8$ $= 2$

• Calculate the magnitude of $\vec{d_2}$ ($||\vec{d_2}||$): $||\vec{d_2}|| = \sqrt{1^2 + \left({3 \over 2}\right)^2 + 2^2}$ $= \sqrt{1 + {9 \over 4} + 4}$ $= \sqrt{{4 \over 4} + {9 \over 4} + {16 \over 4}}$ $= \sqrt{{29 \over 4}} = {{\sqrt{29}} \over 2}$

• Calculate the magnitude of $\vec{d_3}$ ($||\vec{d_3}||$): $||\vec{d_3}|| = \sqrt{(-3)^2 + (-2)^2 + 4^2}$ $= \sqrt{9 + 4 + 16}$ $= \sqrt{29}$

• Apply the angle formula $\cos \theta = {{|\vec{d_2} \cdot \vec{d_3}|} \over {||\vec{d_2}|| \cdot ||\vec{d_3}||}}$: $\cos \theta = {{|2|} \over {\left({{\sqrt{29}} \over 2}\right) (\sqrt{29})}}$ $\cos \theta = {{2} \over {{29} \over 2}}$ $\cos \theta = {{2 \times 2} \over 29}$ $\cos \theta = {{4} \over {29}}$

• Therefore, the angle $\theta$ between lines $l_2$ and $l_3$ is: $\theta = {\cos ^{ - 1}}\left( {{4 \over {29}}} \right)$

Step 4: Compare with Options and Final Answer

We found the angle between lines $l_2$ and $l_3$ to be $\theta = {\cos ^{ - 1}}\left( {{4 \over {29}}} \right)$. Let's check the given options:

(A) ${\cos ^{ - 1}}\left( {{{29} \over 4}} \right)$ (B) ${\sec ^{ - 1}}\left( {{{29} \over 4}} \right)$ (C) ${\cos ^{ - 1}}\left( {{2 \over {29}}} \right)$ (D) ${\cos ^{ - 1}}\left( {{2 \over {\sqrt {29} }}} \right)$

Our calculated value is $\cos \theta = 4/29$. Recall that $\sec \theta = 1/\cos \theta$. So, $\sec \theta = 1 / (4/29) = 29/4$. Therefore, $\theta = {\sec ^{ - 1}}\left( {{{29} \over 4}} \right)$.

This precisely matches option (B). Note that option (A) is mathematically impossible for a real angle since the cosine of any real angle must be between -1 and 1, and $29/4 > 1$.

3. Common Mistakes & Tips

• Always Standardize: The most frequent error is failing to convert line equations to their standard symmetric form ($x-x_1 / a = y-y_1 / b = z-z_1 / c$) before identifying direction ratios. Always ensure the coefficients of $x, y, z$ in the numerator are $+1$. If they are not (e.g., $1-x$ or $2y+3$), factor out the coefficient and adjust the denominator.
• Constant Coordinate: If a coordinate is constant (e.g., $z=k$), its corresponding direction ratio is $0$. This implies the line is perpendicular to that axis (e.g., $z=k$ means the line is perpendicular to the $z$-axis and parallel to the $xy$-plane).
• Acute Angle: The formula for $\cos \theta$ includes an absolute value in the numerator, ensuring that the calculated angle is the acute angle ($0 \le \theta \le \pi/2$).

4. Summary

This problem required a systematic approach to 3D geometry. First, we correctly converted all given line equations into their standard symmetric form to extract their direction vectors. Then, we utilized the perpendicularity condition for lines $l_1$ and $l_2$ to determine the unknown parameter $\alpha$. Finally, with all direction vectors known, we applied the formula for the angle between two lines to find the desired angle between $l_2$ and $l_3$. The calculated cosine of the angle was $4/29$, which corresponds to an angle of $\sec^{-1}(29/4)$.

The final answer is $\boxed{\text{(B)}}$.