1. Key Concepts and Formulas

• Angle Between a Line and a Plane: The angle $\theta$ between a line with direction vector $\vec{b} = (b_1, b_2, b_3)$ and a plane with normal vector $\vec{n} = (n_1, n_2, n_3)$ is given by the formula: $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$ where $\vec{b} \cdot \vec{n}$ is the dot product of the vectors, and $|\vec{b}|$ and $|\vec{n}|$ are their magnitudes. The absolute value ensures that $\sin \theta$ is positive, as the angle between a line and a plane is conventionally taken to be acute ($0^\circ \le \theta \le 90^\circ$).
• Direction Vector of a Line: For a line in symmetric form $\frac{x - x_0}{b_1} = \frac{y - y_0}{b_2} = \frac{z - z_0}{b_3}$, its direction vector is $\vec{b} = (b_1, b_2, b_3)$.
• Normal Vector of a Plane: For a plane in Cartesian form $Ax + By + Cz = D$, its normal vector is $\vec{n} = (A, B, C)$.

2. Step-by-Step Solution

Step 1: Identify the Direction Vector of the Line and the Normal Vector of the Plane. We begin by extracting the necessary vectors from the given equations.

• For the line: $x = {{y - 1} \over 2} = {{z - 3} \over \lambda }$ To match the standard symmetric form $\frac{x - x_0}{b_1} = \frac{y - y_0}{b_2} = \frac{z - z_0}{b_3}$, we can write the given line as $\frac{x - 0}{1} = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$. Thus, the direction vector of the line is $\vec{b} = (1, 2, \lambda)$.

• For the plane: $x+2y+3z=4$ Comparing this with the standard Cartesian form $Ax + By + Cz = D$, we identify the coefficients of $x, y, z$. Thus, the normal vector of the plane is $\vec{n} = (1, 2, 3)$.

Step 2: Calculate the Magnitudes of the Vectors and their Dot Product. These values are essential components of the angle formula.

• Magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2} = \sqrt{1^2 + 2^2 + \lambda^2} = \sqrt{1 + 4 + \lambda^2} = \sqrt{5 + \lambda^2}$

• Magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{n_1^2 + n_2^2 + n_3^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

• Dot product $\vec{b} \cdot \vec{n}$: $\vec{b} \cdot \vec{n} = (1)(1) + (2)(2) + (\lambda)(3) = 1 + 4 + 3\lambda = 5 + 3\lambda$

Step 3: Determine the Value of $\sin \theta$ from the Given Angle. The problem states that the angle between the line and the plane is $\theta = {\cos ^{ - 1}}\left( {\sqrt {{5 \over {14}}} } \right)$. This means $\cos \theta = \sqrt{{5 \over {14}}}$. We need $\sin \theta$ for our formula. Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$:

• $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\sqrt{{5 \over {14}}}\right)^2 = 1 - {5 \over {14}} = {{14 - 5} \over {14}} = {9 \over {14}}$
• Since $\theta$ is an acute angle between a line and a plane, $\sin \theta$ must be positive. $\sin \theta = \sqrt{{9 \over {14}}} = {3 \over {\sqrt{14}}}$

Step 4: Apply the Angle Formula and Solve for $\lambda$. Now, substitute all the calculated values into the formula $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$: ${3 \over {\sqrt{14}}} = \frac{|5 + 3\lambda|}{\left(\sqrt{5 + \lambda^2}\right) \left(\sqrt{14}\right)}$ We can cancel $\sqrt{14}$ from both sides: $3 = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2}}$ To eliminate the absolute value and the square root, square both sides of the equation: $3^2 = \left( \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2}} \right)^2$ $9 = \frac{(5 + 3\lambda)^2}{5 + \lambda^2}$ Cross-multiply and expand the terms: $9(5 + \lambda^2) = (5 + 3\lambda)^2$ $45 + 9\lambda^2 = 25 + 2(5)(3\lambda) + (3\lambda)^2$ $45 + 9\lambda^2 = 25 + 30\lambda + 9\lambda^2$ Cancel $9\lambda^2$ from both sides: $45 = 25 + 30\lambda$ Rearrange the equation to solve for $\lambda$: $45 - 25 = 30\lambda$ $20 = 30\lambda$ $\lambda = {20 \over 30}$ $\lambda = {2 \over 3}$

3. Common Mistakes & Tips

• Incorrect Formula: A very common error is to use $\cos \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$ for the angle between a line and a plane. This formula gives the angle between the line's direction vector and the plane's normal vector, which is complementary to the angle between the line and the plane. Always use $\sin \theta$ for the angle between a line and a plane.
• Vector Identification: Ensure the line equation is in standard symmetric form $\frac{x-x_0}{b_1} = \frac{y-y_0}{b_2} = \frac{z-z_0}{b_3}$ to correctly identify the direction vector $(b_1, b_2, b_3)$. If a term is just $x$, it implies $\frac{x-0}{1}$.
• Algebraic Errors: Be meticulous with squaring both sides, expanding quadratic terms, and simplifying the resulting equation to avoid calculation mistakes. Don't forget the absolute value in the numerator of the formula, as the angle between a line and a plane is conventionally positive.

4. Summary

This problem required finding an unknown parameter $\lambda$ by utilizing the formula for the angle between a line and a plane. The process involved identifying the direction vector of the line and the normal vector of the plane, calculating their magnitudes and dot product, and then substituting these values, along with the given angle (converted to $\sin \theta$), into the formula. Solving the resulting algebraic equation yielded the value of $\lambda$.

5. Final Answer

The final answer is $\boxed{{3 \over 2}}$, which corresponds to option (A).