1. Key Concepts and Formulas

   • Equation of a Plane Containing Two Intersecting Lines: If two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ intersect, they lie in a unique plane. The equation of this plane can be found using the common point ($\vec{a}_1$ or $\vec{a}_2$) and the normal vector to the plane. The normal vector $\vec{n}$ is perpendicular to both direction vectors $\vec{b}_1$ and $\vec{b}_2$, and thus can be found by their cross product: $\vec{n} = \vec{b}_1 \times \vec{b}_2$. The equation of the plane in vector form is $\vec{n} \cdot (\vec{r} - \vec{a}) = 0$, where $\vec{a}$ is the position vector of the common point. In Cartesian form, if $\vec{n} = (A, B, C)$ and $\vec{a} = (x_1, y_1, z_1)$, the equation is $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.

   • Distance from a Point to a Plane: The perpendicular distance $d$ from a point $P(x_0, y_0, z_0)$ to a plane given by the Cartesian equation $Ax + By + Cz + D = 0$ is calculated using the formula: $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

2. Step-by-Step Solution

   Step 1: Identify the common point and direction vectors of the lines. The given lines are: Line 1: $\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$ Line 2: $\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$

   • Reasoning: Both lines pass through the point whose position vector is $\widehat i + \widehat j$. This means they intersect at the point $A(1, 1, 0)$. This point lies on the plane.
   • The direction vector of Line 1 is $\vec{b}_1 = \widehat i + 2\widehat j - \widehat k$, which can be written as $(1, 2, -1)$.
   • The direction vector of Line 2 is $\vec{b}_2 = - \widehat i + \widehat j - 2\widehat k$, which can be written as $(-1, 1, -2)$.
   • Why: We need a point on the plane and a normal vector to define the plane's equation. The common point serves as the point, and the direction vectors are used to find the normal.

   Step 2: Calculate the normal vector to the plane.

   • Reasoning: Since both lines lie in the plane, their direction vectors are parallel to the plane. The cross product of these two direction vectors will yield a vector perpendicular (normal) to the plane.
   • Calculation: $\vec{n} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix}$ $\vec{n} = \widehat i ((2)(-2) - (-1)(1)) - \widehat j ((1)(-2) - (-1)(-1)) + \widehat k ((1)(1) - (2)(-1))$ $\vec{n} = \widehat i (-4 + 1) - \widehat j (-2 - 1) + \widehat k (1 + 2)$ $\vec{n} = -3\widehat i + 3\widehat j + 3\widehat k$
   • Why: We can use any scalar multiple of $\vec{n}$ as the normal vector. For simplicity, we can divide by 3: $\vec{n}' = -\widehat i + \widehat j + \widehat k$ So, the coefficients of the plane equation will be $(A, B, C) = (-1, 1, 1)$.

   Step 3: Formulate the equation of the plane.

   • Reasoning: We have a point on the plane $A(1, 1, 0)$ and a normal vector $\vec{n}' = (-1, 1, 1)$. We can use the Cartesian form of the plane equation $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.
   • Calculation: $-1(x - 1) + 1(y - 1) + 1(z - 0) = 0$ $-x + 1 + y - 1 + z = 0$ $-x + y + z = 0$ This can also be written as $x - y - z = 0$.
   • Why: This equation represents the plane that contains the two given lines. We need this equation to calculate the distance from the given point.

   Step 4: Calculate the perpendicular distance from the point (2, 1, 4) to the plane.

   • Reasoning: We will use the formula for the distance from a point to a plane.
   • Given: Point $P(x_0, y_0, z_0) = (2, 1, 4)$.
   • Plane Equation: $x - y - z = 0$, so $A = 1, B = -1, C = -1, D = 0$.
   • Calculation: $d = \frac{|A x_0 + B y_0 + C z_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ $d = \frac{|1(2) + (-1)(1) + (-1)(4) + 0|}{\sqrt{1^2 + (-1)^2 + (-1)^2}}$ $d = \frac{|2 - 1 - 4|}{\sqrt{1 + 1 + 1}}$ $d = \frac{|-3|}{\sqrt{3}}$ $d = \frac{3}{\sqrt{3}}$ To rationalize the denominator: $d = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}$
   • Why: This is the final step to directly answer the question asked.

3. Common Mistakes & Tips

   • Cross Product Errors: Be very careful with the signs and calculations when computing the cross product. A small error here will propagate through the entire problem.
   • Sign in Distance Formula: Remember the absolute value in the numerator of the distance formula. Distance is always non-negative.
   • Point on Plane: Ensure the chosen point for the plane equation actually lies on the plane (e.g., the common point of the lines).
   • Normal Vector Scaling: Any non-zero scalar multiple of a normal vector is also a normal vector. Simplifying the normal vector (e.g., dividing by a common factor) can make calculations easier, as done in Step 2.

4. Summary

   To find the perpendicular distance from the point (2, 1, 4) to the plane containing the given lines, we first identified the common point (1, 1, 0) and the direction vectors of the lines. We then calculated the normal vector to the plane using the cross product of the direction vectors, which was $(-3, 3, 3)$ or simplified to $(-1, 1, 1)$. Using this normal vector and the common point, we found the equation of the plane to be $x - y - z = 0$. Finally, we applied the formula for the distance from a point to a plane, using the point (2, 1, 4) and the plane $x - y - z = 0$, which yielded a distance of $\sqrt{3}$.

5. Final Answer

The final answer is $\boxed{\sqrt{3}}$, which corresponds to option (D).