The unit vector normal to the plane `x + 2y +3z-6 =0` is `(1)/(sqrt(14)) hati + (2)/(sqrt(14))hatj + (3)/(sqrt(14))hatk.`

Answer

To determine whether the given unit vector normal to the plane \(x + 2y + 3z - 6 = 0\) is correct, we will follow these steps:

### Step 1: Identify the normal vector from the plane equation
The general form of a plane equation is given by:
\[ Ax + By + Cz = D \]
From the equation \(x + 2y + 3z - 6 = 0\), we can rewrite it as:
\[ x + 2y + 3z = 6 \]
Here, we can identify \(A = 1\), \(B = 2\), and \(C = 3\). Therefore, the normal vector \(\mathbf{n}\) to the plane is:
\[ \mathbf{n} = Ai + Bj + Ck = 1i + 2j + 3k \]

### Step 2: Calculate the magnitude of the normal vector
The magnitude of the normal vector \(\mathbf{n}\) is calculated using the formula:
\[ |\mathbf{n}| = \sqrt{A^2 + B^2 + C^2} \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ |\mathbf{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 3: Find the unit vector normal to the plane
The unit vector \(\mathbf{n}_{\text{unit}}\) in the direction of the normal vector is given by:
\[ \mathbf{n}_{\text{unit}} = \frac{\mathbf{n}}{|\mathbf{n}|} \]
Substituting the values we found:
\[ \mathbf{n}_{\text{unit}} = \frac{1i + 2j + 3k}{\sqrt{14}} = \frac{1}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j + \frac{3}{\sqrt{14}}k \]

### Conclusion
Thus, the unit vector normal to the plane \(x + 2y + 3z - 6 = 0\) is:
\[ \frac{1}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j + \frac{3}{\sqrt{14}}k \]
This matches the given unit vector, confirming that the statement is **true**.