If the plane containing the line `(x-3)/2 = (y+2)/(-1) = (z-1)/3` and also containing its projection on the plane `2x+3y-z=5` contains which one of the following points? (a) `(2,2,0)` (b) `(2,0,-2)` (c) `(0,-2,2)` (d) `(-2,2,2)`

Options

A. (2,2,0)
B. (-2,2,2)
C. (0,-2,2)
D. (2,0,-2)

Explanation

To solve the problem, we need to determine the equation of the required plane that contains the given line and its projection onto the plane \(2x + 3y - z = 5\). We will then check which of the given points lies on this plane.

### Step 1: Identify the line and its direction vector
The line is given in the symmetric form:
\[ \frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3} \]
From this, we can extract the direction vector of the line:
\[ \mathbf{b} = \langle 2, -1, 3 \rangle \]
And a point on the line, which we can take as \(A(3, -2, 1)\).

### Step 2: Find the normal vector of the given plane
The equation of the plane is:
\[ 2x + 3y - z = 5 \]
The normal vector \(\mathbf{n}\) of this plane can be directly read from the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n} = \langle 2, 3, -1 \rangle \]

### Step 3: Find the normal vector of the required plane
The required plane contains the line and its projection onto the given plane. To find the normal vector of the required plane, we need to take the cross product of the direction vector of the line and the normal vector of the given plane:
\[ \mathbf{n}_{\text{required}} = \mathbf{b} \times \mathbf{n} \]

Calculating the cross product:
\[ \mathbf{b} = \langle 2, -1, 3 \rangle, \quad \mathbf{n} = \langle 2, 3, -1 \rangle \]
\[ \mathbf{n}_{\text{required}} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 2 & 3 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i}((-1)(-1) - (3)(3)) - \mathbf{j}((2)(-1) - (3)(2)) + \mathbf{k}((2)(3) - (-1)(2)) \]
\[ = \mathbf{i}(1 - 9) - \mathbf{j}(-2 - 6) + \mathbf{k}(6 + 2) \]
\[ = \mathbf{i}(-8) + \mathbf{j}(8) + \mathbf{k}(8) \]
Thus, we have:
\[ \mathbf{n}_{\text{required}} = \langle -8, 8, 8 \rangle \]

### Step 4: Equation of the required plane
The equation of a plane can be expressed as:
\[ -8(x - 3) + 8(y + 2) + 8(z - 1) = 0 \]
Expanding this:
\[ -8x + 24 + 8y + 16 + 8z - 8 = 0 \]
Combining like terms:
\[ -8x + 8y + 8z + 32 = 0 \]
Dividing by -8:
\[ x - y - z = 4 \]

### Step 5: Check which point satisfies the plane equation
Now we will check which of the given points satisfies the equation \(x - y - z = 4\).

1. For point \( (2, 2, 0) \):
\[ 2 - 2 - 0 = 0 \quad (\text{not } 4) \]
2. For point \( (2, 0, -2) \):
\[ 2 - 0 - (-2) = 2 + 2 = 4 \quad (\text{this works!}) \]
3. For point \( (0, -2, 2) \):
\[ 0 - (-2) - 2 = 0 + 2 - 2 = 0 \quad (\text{not } 4) \]
4. For point \( (-2, 2, 2) \):
\[ -2 - 2 - 2 = -6 \quad (\text{not } 4) \]

### Conclusion
The only point that satisfies the equation of the required plane is:
\[ \boxed{(2, 0, -2)} \]