To find the distance of the point \((\frac{3\sqrt{3}}{2}, 3\sqrt{3}, 3)\) from the plane \(P\) that passes through the point \((1, 1, 1)\) and is parallel to the vectors \(\vec{a} = -\hat{i} + \hat{j}\) and \(\vec{b} = \hat{i} - \hat{k}\), we will follow these steps:
### Step 1: Find the normal vector of the plane
The normal vector \(\vec{n}\) of the plane can be found by taking the cross product of the vectors \(\vec{a}\) and \(\vec{b}\).
\[
\vec{a} = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}
\]
The cross product \(\vec{a} \times \vec{b}\) is calculated as follows:
\[
\vec{n} = \vec{a} \times \vec{b} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & 0 \\
1 & 0 & -1
\end{vmatrix}
\]
Calculating the determinant:
\[
\vec{n} = \hat{i} \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 0 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 1 & 0 \end{vmatrix}
\]
\[
= \hat{i} (1 \cdot -1 - 0 \cdot 0) - \hat{j} (-1 \cdot -1 - 0 \cdot 1) + \hat{k} (-1 \cdot 0 - 1 \cdot 1)
\]
\[
= -\hat{i} - \hat{j} - \hat{k}
\]
Thus, the normal vector is \(\vec{n} = -\hat{i} - \hat{j} - \hat{k}\).
### Step 2: Write the equation of the plane
The equation of the plane can be written using the point-normal form:
\[
\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}
\]
Where \(\vec{r} = (x, y, z)\) and \(\vec{a} = (1, 1, 1)\).
Calculating \(\vec{a} \cdot \vec{n}\):
\[
\vec{a} \cdot \vec{n} = (1)(-1) + (1)(-1) + (1)(-1) = -3
\]
Thus, the equation of the plane becomes:
\[
-x - y - z = -3 \quad \text{or} \quad x + y + z = 3
\]
### Step 3: Find the distance from the point to the plane
The distance \(D\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is given by:
\[
D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
For our plane \(x + y + z - 3 = 0\), we have \(A = 1\), \(B = 1\), \(C = 1\), and \(D = -3\). The point is \((\frac{3\sqrt{3}}{2}, 3\sqrt{3}, 3)\).
Calculating the numerator:
\[
D = \frac{|1 \cdot \frac{3\sqrt{3}}{2} + 1 \cdot 3\sqrt{3} + 1 \cdot 3 - 3|}{\sqrt{1^2 + 1^2 + 1^2}}
\]
Calculating:
\[
= \frac{|\frac{3\sqrt{3}}{2} + 3\sqrt{3} + 3 - 3|}{\sqrt{3}} = \frac{|\frac{3\sqrt{3}}{2} + 3\sqrt{3}|}{\sqrt{3}} = \frac{|\frac{3\sqrt{3}}{2} + \frac{6\sqrt{3}}{2}|}{\sqrt{3}} = \frac{|\frac{9\sqrt{3}}{2}|}{\sqrt{3}} = \frac{9}{2}
\]
### Final Answer
Thus, the distance of the point from the plane is:
\[
D = \frac{9}{2} = 4.5
\]
