1. Key Concepts and Formulas

• Direction Vector of a Line: For a line given in symmetric form ${{x - x_1} \over a} = {{y - y_1} \over b} = {{z - z_1} \over c}$, its direction vector is $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$. These $(a,b,c)$ are known as the direction ratios of the line.
• Normal Vector of a Plane: For a plane given in general form $Ax + By + Cz + D = 0$, its normal vector is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. The coefficients $(A,B,C)$ are the direction ratios of the normal to the plane.
• Angle between a Line and a Plane: If $\theta$ is the angle between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$, then the formula relating them is: $\sin \theta = \left| {{{\vec{b} \cdot \vec{n}} \over {|\vec{b}| |\vec{n}|}}} \right|$ Here, $\vec{b} \cdot \vec{n}$ 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 $\theta$ is conventionally taken as an acute angle ($0 \le \theta \le \pi/2$).

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 essential vector information from the given equations.

The equation of the line is given as: ${{x + 1} \over 2} = {{y - 2} \over 1} = {{z - 3} \over { - 2}}$ By comparing this with the standard symmetric form of a line ${{x - x_1} \over a} = {{y - y_1} \over b} = {{z - z_1} \over c}$, we can identify the direction ratios $(a, b, c)$ of the line as $(2, 1, -2)$. Therefore, the direction vector of the line, $\vec{b}$, is: $\vec{b} = 2\hat{i} + 1\hat{j} - 2\hat{k}$

The equation of the plane is given as: $x - 2y - kz = 3$ To find the normal vector, we rewrite this in the standard general form $Ax + By + Cz + D = 0$: $1x - 2y - kz - 3 = 0$ By comparing this with $Ax + By + Cz + D = 0$, we identify the coefficients $(A, B, C)$ as $(1, -2, -k)$. These coefficients represent the direction ratios of the normal to the plane. Therefore, the normal vector to the plane, $\vec{n}$, is: $\vec{n} = 1\hat{i} - 2\hat{j} - k\hat{k}$

Step 2: Calculate the Magnitudes of the Vectors and their Dot Product. Next, we compute the magnitudes of the direction vector of the line and the normal vector of the plane, as well as their dot product. These values are necessary for the angle formula.

Magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{1^2 + (-2)^2 + (-k)^2} = \sqrt{1 + 4 + k^2} = \sqrt{5 + k^2}$

Dot product $\vec{b} \cdot \vec{n}$: $\vec{b} \cdot \vec{n} = (2)(1) + (1)(-2) + (-2)(-k) = 2 - 2 + 2k = 2k$

Step 3: Use the Formula for the Angle between a Line and a Plane. Let $\theta$ be the angle between the line and the plane. We apply the formula stated in the Key Concepts: $\sin \theta = \left| {{{\vec{b} \cdot \vec{n}} \over {|\vec{b}| |\vec{n}|}}} \right|$ Substitute the values calculated in Step 2 into this formula: $\sin \theta = \left| {{{2k} \over {3\sqrt{5 + k^2}}}} \right|$ Since the angle $\theta$ between a line and a plane is conventionally taken to be acute ($0 \le \theta \le \pi/2$), $\sin \theta$ must be non-negative. Thus, we write: $\sin \theta = {{|2k|} \over {3\sqrt{5 + k^2}}} \quad \quad \text{(Equation 1)}$

Step 4: Determine $\sin \theta$ from the Given Angle. The problem provides the angle between the line and the plane as ${\cos ^{ - 1}}\left( {{{2\sqrt 2 } \over 3}} \right)$. This means that $\cos \theta = {{2\sqrt 2 } \over 3}$. To find $\sin \theta$, we use the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$. Since $\theta$ is an acute angle, $\sin \theta$ will be positive. $\sin^2 \theta = 1 - \cos^2 \theta$ $\sin^2 \theta = 1 - \left( {{{2\sqrt 2 } \over 3}} \right)^2$ $\sin^2 \theta = 1 - {{4 \cdot 2} \over 9}$ $\sin^2 \theta = 1 - {8 \over 9}$ $\sin^2 \theta = {1 \over 9}$ Taking the positive square root to find $\sin \theta$: $\sin \theta = \sqrt{{1 \over 9}} = {1 \over 3} \quad \quad \text{(Equation 2)}$

Step 5: Equate the Expressions for $\sin \theta$ and Solve for $k$. Now, we have two expressions for $\sin \theta$ (Equation 1 and Equation 2). We equate them to solve for $k$: ${{|2k|} \over {3\sqrt{5 + k^2}}} = {1 \over 3}$ To simplify, multiply both sides of the equation by 3: ${{|2k|} \over {\sqrt{5 + k^2}}} = 1$ $|2k| = \sqrt{5 + k^2}$ To eliminate the absolute value and the square root, we square both sides of the equation: $(|2k|)^2 = (\sqrt{5 + k^2})^2$ $4k^2 = 5 + k^2$ Now, we solve this algebraic equation for $k$: $4k^2 - k^2 = 5$ $3k^2 = 5$ $k^2 = {5 \over 3}$ Taking the square root of both sides gives us the possible values for $k$: $k = \pm \sqrt{{5 \over 3}}$ The question asks for "a value of k". Among the given options, $\sqrt{{5 \over 3}}$ is present.

3. Common Mistakes & Tips

• Confusing Angle between Line and Plane vs. Line and Normal: A common error is to directly use the dot product formula for the angle between two vectors ($\cos \phi = {{|\vec{b} \cdot \vec{n}|} \over {|\vec{b}| |\vec{n}|}}$). This formula gives the angle $\phi$ between the line and the normal to the plane. The angle $\theta$ between the line and the plane itself is complementary to $\phi$, i.e., $\theta = 90^\circ - \phi$. Therefore, $\sin \theta = \cos \phi$. The formula $\sin \theta = \left| {{{\vec{b} \cdot \vec{n}} \over {|\vec{b}| |\vec{n}|}}} \right|$ already incorporates this relationship.
• Forgetting Absolute Value: The angle between a line and a plane is always considered to be acute ($0 \le \theta \le \pi/2$), so its sine value must be non-negative. Always remember to include the absolute value in the numerator of the formula, $\left| \vec{b} \cdot \vec{n} \right|$, to ensure $\sin \theta \ge 0$.
• Algebraic Precision: Be careful with squaring terms and handling square roots. Ensure $(2k)^2$ is correctly evaluated as $4k^2$ and not $2k^2$. Squaring both sides of an equation involving absolute values or square roots is a standard technique, but always check for extraneous solutions if the original equation was non-linear. In this case, both positive and negative values of $k$ satisfy the squared equation, and we pick the one from the options.

4. Summary

This problem required us to find a missing parameter 'k' by utilizing the formula for the angle between a line and a plane in 3D geometry. We started by extracting the direction vector of the line and the normal vector of the plane from their respective equations. Then, we calculated their magnitudes and dot product. Substituting these into the angle formula gave us an expression for $\sin \theta$ in terms of $k$. Separately, we used the given $\cos \theta$ value to determine the numerical value of $\sin \theta$. By equating these two expressions for $\sin \theta$, we formed an equation that was solved for $k^2$, yielding $k = \pm \sqrt{{5 \over 3}}$.

5. Final Answer

The value of $k$ is $\pm \sqrt{{5 \over 3}}$. From the given options, a value of $k$ is $\sqrt{{5 \over 3}}$. The final answer is $\boxed{\sqrt{{5 \over 3}}}$ which corresponds to option (D).