Show that the lines whose vector equation is `vec(r) = (hat(i) + hat(j) ) + lambda (2 hat(i) + hat(j) + 4 hat(k))` is parallel to the plane whose vector equation is `vec(r). (hat(i) + 2 hat(j) - hat(k) ) = 3`, and find the distance between them. Also, state whether the line lies in the plane.

Answer

To solve the problem, we need to show that the given line is parallel to the given plane, find the distance between them, and determine if the line lies in the plane.

### Step 1: Identify the Vector Equation of the Line
The vector equation of the line is given as:
\[ \vec{r} = (\hat{i} + \hat{j}) + \lambda (2 \hat{i} + \hat{j} + 4 \hat{k}) \]
From this equation, we can identify a point on the line and the direction ratios.

**Point on the line**: When \(\lambda = 0\), we have the point \( (1, 1, 0) \).

**Direction ratios of the line**: The coefficients of \(\lambda\) give us the direction ratios:
\[ \text{Direction ratios} = (2, 1, 4) \]

### Step 2: Identify the Equation of the Plane
The vector equation of the plane is given as:
\[ \vec{r} \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 3 \]
We can express this in Cartesian form. Let \(\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}\). Then,
\[ x + 2y - z = 3 \]
This can be rearranged to:
\[ x + 2y - z - 3 = 0 \]

### Step 3: Find the Normal Vector of the Plane
The normal vector of the plane can be derived from the coefficients of \(x\), \(y\), and \(z\) in the equation:
\[ \text{Normal vector} = (1, 2, -1) \]

### Step 4: Check if the Line is Parallel to the Plane
For the line to be parallel to the plane, the direction ratios of the line must be perpendicular to the normal vector of the plane. This can be checked using the dot product:
\[ \text{Direction ratios of the line} = (2, 1, 4) \]
\[ \text{Normal vector of the plane} = (1, 2, -1) \]
Calculating the dot product:
\[ (2, 1, 4) \cdot (1, 2, -1) = 2 \cdot 1 + 1 \cdot 2 + 4 \cdot (-1) = 2 + 2 - 4 = 0 \]
Since the dot product is zero, the line is parallel to the plane.

### Step 5: Find the Distance from the Line to the Plane
To find the distance from the line to the plane, we can use the formula for the distance \(D\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\):
\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Using the point \((1, 1, 0)\) from the line and the plane equation \(x + 2y - z - 3 = 0\) (where \(A=1\), \(B=2\), \(C=-1\), and \(D=-3\)):
\[ D = \frac{|1 \cdot 1 + 2 \cdot 1 - 1 \cdot 0 - 3|}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{|1 + 2 - 3|}{\sqrt{1 + 4 + 1}} = \frac{|0|}{\sqrt{6}} = 0 \]

### Step 6: Conclusion
Since the distance is \(0\), it indicates that the line lies in the plane.

### Final Answer
1. The line is parallel to the plane.
2. The distance between the line and the plane is \(0\).
3. The line lies in the plane.