A plane P passes through the point `(1, 1,1)` and is parallel to the vectors `veca=-hati+hatj and vecb=hati-hatk`. The distance of the point `((3sqrt3)/(2), 3sqrt3, 3)` from the plane is equal to

Options

A. `sqrt3" units"`
B. `(9)/(2)" units"`
C. `3sqrt3" units"`
D. 3 units

Answer

To find the distance of the point \(\left(\frac{3\sqrt{3}}{2}, 3\sqrt{3}, 3\right)\) 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 using the cross product of the vectors \(\vec{a}\) and \(\vec{b}\).

\[ \vec{a} = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \]

Calculating the cross product \(\vec{a} \times \vec{b}\):

\[ \vec{n} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 1 & 0 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{n} = \hat{i} \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 0 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 1 & 0 \end{vmatrix} \]

Calculating each determinant:

\[ = \hat{i}(1 \cdot -1 - 0 \cdot 0) - \hat{j}(-1 \cdot -1 - 0 \cdot 1) + \hat{k}(-1 \cdot 0 - 1 \cdot 1) \]
\[ = -\hat{i} - \hat{j} - \hat{k} \]

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 in the form \(\vec{r} \cdot \vec{n} = d\) can be derived from the point \((1, 1, 1)\):

Using the normal vector \(\vec{n} = (-1, -1, -1)\) and the point \((1, 1, 1)\):

\[ \vec{r} \cdot (-1, -1, -1) = d \]

Calculating \(d\):

\[ (1, 1, 1) \cdot (-1, -1, -1) = -1 - 1 - 1 = -3 \]

Thus, the equation of the plane is:

\[ \vec{r} \cdot (-1, -1, -1) = -3 \]

### 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 given by \(\vec{r} \cdot \vec{n} = d\) is given by:

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

Where \(\vec{p} = \left(\frac{3\sqrt{3}}{2}, 3\sqrt{3}, 3\right)\).

Calculating \(\vec{n} \cdot \vec{p}\):

\[ \vec{n} \cdot \vec{p} = (-1, -1, -1) \cdot \left(\frac{3\sqrt{3}}{2}, 3\sqrt{3}, 3\right) \]
\[ = -\frac{3\sqrt{3}}{2} - 3\sqrt{3} - 3 \]
\[ = -\frac{3\sqrt{3}}{2} - \frac{6\sqrt{3}}{2} - \frac{6}{2} = -\frac{3\sqrt{3} + 6\sqrt{3} + 6}{2} = -\frac{9\sqrt{3} + 6}{2} \]

Thus,

\[ D = \frac{\left| -\frac{9\sqrt{3} + 6}{2} + 3 \right|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\left| -\frac{9\sqrt{3} + 6}{2} + \frac{6}{2} \right|}{\sqrt{3}} = \frac{\left| -\frac{9\sqrt{3}}{2} \right|}{\sqrt{3}} = \frac{9\sqrt{3}/2}{\sqrt{3}} = \frac{9}{2} \]

### Final Answer
The distance of the point from the plane is:

\[ \boxed{\frac{9}{2}} \]