Find the the distance of a point (2,5, -3) from the plane `vec(r).(6 hati - 3 hatj + 2 hatk) = 4.`

Answer

To find the distance of the point \( (2, 5, -3) \) from the plane given by the equation \( \vec{r} \cdot (6 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4 \), we can follow these steps:

### Step 1: Convert the Plane Equation to Cartesian Form
The given equation of the plane in vector form is:
\[ \vec{r} \cdot (6 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4 \]
This can be expressed in Cartesian form as:
\[ 6x - 3y + 2z = 4 \]

### Step 2: Identify the Coefficients
From the Cartesian equation \( 6x - 3y + 2z = 4 \), we can identify the coefficients:
- \( A = 6 \)
- \( B = -3 \)
- \( C = 2 \)
- \( D = 4 \)

### Step 3: Use the Distance Formula
The formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz = D \) is given by:
\[ D = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here, the point is \( (x_1, y_1, z_1) = (2, 5, -3) \).

### Step 4: Substitute the Values into the Formula
Substituting the values into the distance formula:
\[ D = \frac{|6(2) + (-3)(5) + 2(-3) - 4|}{\sqrt{6^2 + (-3)^2 + 2^2}} \]

### Step 5: Calculate the Numerator
Calculating the numerator:
\[ 6(2) = 12 \]
\[ -3(5) = -15 \]
\[ 2(-3) = -6 \]
Now, combine these:
\[ 12 - 15 - 6 - 4 = 12 - 15 - 6 - 4 = -13 \]
Taking the absolute value:
\[ |-13| = 13 \]

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

### Step 7: Final Calculation of Distance
Now, substituting back into the distance formula:
\[ D = \frac{13}{7} \]

### Conclusion
Thus, the distance of the point \( (2, 5, -3) \) from the plane is:
\[ \frac{13}{7} \text{ units} \]

---