Find the perpendicular distance from the point `(2hati+hatj-hatk)` to the plane`vecr.(i-2hatj+4hatk) = 3`.

Answer

To find the perpendicular distance from the point \( \vec{A} = 2\hat{i} + \hat{j} - \hat{k} \) to the plane defined by the equation \( \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 3 \), we can use the formula for the distance \( P \) from a point to a plane:

\[ P = \frac{|\vec{A} \cdot \vec{n} - d|}{|\vec{n}|} \]

where:
- \( \vec{A} \) is the position vector of the point,
- \( \vec{n} \) is the normal vector to the plane,
- \( d \) is the constant from the plane equation.

### Step 1: Identify the vectors
The normal vector \( \vec{n} \) can be extracted from the plane equation. From the equation \( \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 3 \), we have:
\[ \vec{n} = \hat{i} - 2\hat{j} + 4\hat{k} \]
and \( d = 3 \).

The position vector of the point is:
\[ \vec{A} = 2\hat{i} + \hat{j} - \hat{k} \]

### Step 2: Calculate the dot product \( \vec{A} \cdot \vec{n} \)
Now, we calculate the dot product \( \vec{A} \cdot \vec{n} \):
\[ \vec{A} \cdot \vec{n} = (2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) \]
Calculating this gives:
\[ = 2 \cdot 1 + 1 \cdot (-2) + (-1) \cdot 4 = 2 - 2 - 4 = -4 \]

### Step 3: Substitute into the distance formula
Now we substitute \( \vec{A} \cdot \vec{n} \) and \( d \) into the distance formula:
\[ P = \frac{|-4 - 3|}{|\vec{n}|} \]
Calculating the numerator:
\[ |-4 - 3| = |-7| = 7 \]

### Step 4: Calculate the magnitude of \( \vec{n} \)
Next, we calculate the magnitude of the normal vector \( \vec{n} \):
\[ |\vec{n}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \]

### Step 5: Final calculation of distance
Now we can find the distance \( P \):
\[ P = \frac{7}{\sqrt{21}} \]

### Conclusion
Thus, the perpendicular distance from the point \( (2, 1, -1) \) to the plane is:
\[ P = \frac{7}{\sqrt{21}} \text{ units} \]