Find the perpendicular distance from the point `(2hati-hatj+4hatk)` to the plane `vecr.(3hati-4hatj+12hatk) = 1`.

Answer

To find the perpendicular distance from the point \( \vec{A} = 2\hat{i} - \hat{j} + 4\hat{k} \) to the plane defined by the equation \( \vec{r} \cdot (3\hat{i} - 4\hat{j} + 12\hat{k}) = 1 \), we can use the formula for the distance \( d \) from a point to a plane:

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

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

### Step 1: Identify the vectors
The normal vector \( \vec{n} \) of the plane can be extracted from the equation of the plane. Here, \( \vec{n} = 3\hat{i} - 4\hat{j} + 12\hat{k} \) and \( d = 1 \).

### Step 2: Calculate the dot product \( \vec{A} \cdot \vec{n} \)
Now, we calculate the dot product \( \vec{A} \cdot \vec{n} \):

\[ \vec{A} = 2\hat{i} - \hat{j} + 4\hat{k} \]

\[ \vec{n} = 3\hat{i} - 4\hat{j} + 12\hat{k} \]

Calculating the dot product:

\[ \vec{A} \cdot \vec{n} = (2)(3) + (-1)(-4) + (4)(12) \]

\[ = 6 + 4 + 48 = 58 \]

### Step 3: Substitute into the distance formula
Now we substitute \( \vec{A} \cdot \vec{n} \) and \( d \) into the distance formula:

\[ d = \frac{|58 - 1|}{|\vec{n}|} \]

\[ = \frac{|57|}{|\vec{n}|} \]

### Step 4: Calculate the magnitude of \( \vec{n} \)
Next, we calculate the magnitude of the normal vector \( \vec{n} \):

\[ |\vec{n}| = \sqrt{(3)^2 + (-4)^2 + (12)^2} \]

\[ = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \]

### Step 5: Final calculation of distance
Now we can find the distance \( d \):

\[ d = \frac{57}{13} \]

Calculating this gives:

\[ d = 4.3846 \text{ (approximately)} \]

### Final Answer
Thus, the perpendicular distance from the point \( (2, -1, 4) \) to the plane is approximately \( 4.38 \) units.