The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines `r=(hati+hatj)+lambda(hati+2hatj-hatk)" and "r=(hati+hatj)+mu(-hati+hatj-2hatk)` is

Explanation

To find the length of the perpendicular drawn from the point \( P(2, 1, 4) \) to the plane containing the given lines, we will follow these steps:

### Step 1: Identify the direction vectors of the lines
The two lines given are:
1. \( \mathbf{r_1} = \mathbf{i} + \mathbf{j} + \lambda(\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \)
2. \( \mathbf{r_2} = \mathbf{i} + \mathbf{j} + \mu(-\mathbf{i} + \mathbf{j} - 2\mathbf{k}) \)

From these equations, we can extract the direction vectors:
- For the first line, the direction vector \( \mathbf{d_1} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} \)
- For the second line, the direction vector \( \mathbf{d_2} = -\mathbf{i} + \mathbf{j} - 2\mathbf{k} \)

### Step 2: Find the normal vector to the plane
The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of the two direction vectors:
\[ \mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} \]

Calculating the cross product:
\[ \mathbf{d_1} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}, \quad \mathbf{d_2} = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix} \]

Using the determinant:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix} \]

Calculating this determinant:
\[ \mathbf{n} = \mathbf{i} \left( 2 \cdot (-2) - (-1) \cdot 1 \right) - \mathbf{j} \left( 1 \cdot (-2) - (-1) \cdot (-1) \right) + \mathbf{k} \left( 1 \cdot 1 - 2 \cdot (-1) \right) \]
\[ = \mathbf{i} (-4 + 1) - \mathbf{j} (-2 - 1) + \mathbf{k} (1 + 2) \]
\[ = -3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} \]

Thus, the normal vector is:
\[ \mathbf{n} = -3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} \]

### Step 3: Write the equation of the plane
Using the point \( (1, 1, 1) \) (from the lines) and the normal vector \( \mathbf{n} \):
\[ -3(x - 1) + 3(y - 1) + 3(z - 1) = 0 \]
Expanding this:
\[ -3x + 3 + 3y - 3 + 3z - 3 = 0 \]
This simplifies to:
\[ -3x + 3y + 3z = 0 \quad \Rightarrow \quad x - y - z = 0 \]

### Step 4: Calculate the distance from point to the plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For our plane \( x - y - z = 0 \):
- \( A = 1, B = -1, C = -1, D = 0 \)
- The point \( P(2, 1, 4) \)

Substituting into the distance formula:
\[ d = \frac{|1(2) - 1(1) - 1(4) + 0|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|2 - 1 - 4|}{\sqrt{1 + 1 + 1}} = \frac{|-3|}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

### Final Answer
The length of the perpendicular drawn from the point \( (2, 1, 4) \) to the plane is:
\[ \sqrt{3} \]