Find the distance of the point (2,3,4) from the plane :
`vec(r) . (3 hati - 6 hatj + 2 hatk ) = - 11.`

Answer

To find the distance of the point \( P(2, 3, 4) \) from the plane given by the equation \( \vec{r} \cdot (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = -11 \), we will first convert the vector form of the plane equation into Cartesian form and then apply the formula for the distance from a point to a plane.

### Step 1: Convert the plane equation to Cartesian form

The vector equation of the plane is given as:
\[ \vec{r} \cdot (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = -11 \]

Let \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \). The dot product can be expressed as:
\[ (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = 3x - 6y + 2z \]

Setting this equal to \(-11\), we have:
\[ 3x - 6y + 2z = -11 \]

Rearranging this gives us the Cartesian form of the plane:
\[ 3x - 6y + 2z + 11 = 0 \]

### Step 2: Identify coefficients for the distance formula

From the equation of the plane \( ax + by + cz + d = 0 \), we identify:
- \( a = 3 \)
- \( b = -6 \)
- \( c = 2 \)
- \( d = 11 \)

### Step 3: Use the distance formula

The formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( ax + by + cz + d = 0 \) is given by:
\[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

Substituting \( (x_1, y_1, z_1) = (2, 3, 4) \):
\[ D = \frac{|3(2) + (-6)(3) + 2(4) + 11|}{\sqrt{3^2 + (-6)^2 + 2^2}} \]

### Step 4: Calculate the numerator

Calculating the numerator:
\[ 3(2) + (-6)(3) + 2(4) + 11 = 6 - 18 + 8 + 11 = 7 \]
Thus, the absolute value is:
\[ |7| = 7 \]

### Step 5: Calculate the denominator

Calculating the denominator:
\[ \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

### Step 6: Calculate the distance

Now substituting back into the distance formula:
\[ D = \frac{7}{7} = 1 \]

### Final Answer

The distance of the point \( (2, 3, 4) \) from the plane is \( \boxed{1} \).