If the volume of tetrahedron `A B C D` is 1 cubic units, where `A(0,1,2),B(-1,2,1)` and `C(1,2,1),` then the locus of point `D` is a. `x+y-z=3` b. `y+z=6` c. `y+z=0` d. `y+z=-3`

Options

A. `x+y-z=0`
B. `y+x=6`
C. `y+z=0`
D. `y+z=-3`

Explanation

To find the locus of point \( D \) given the volume of tetrahedron \( ABCD \) is 1 cubic unit, we will follow these steps:

### Step 1: Understand the Volume Formula
The volume \( V \) of a tetrahedron formed by points \( A, B, C, D \) can be calculated using the formula:
\[ V = \frac{1}{6} | \vec{AB} \cdot (\vec{AC} \times \vec{AD}) | \]
where \( \vec{AB}, \vec{AC}, \vec{AD} \) are vectors from point \( A \) to points \( B, C, D \) respectively.

### Step 2: Calculate Vectors \( \vec{AB}, \vec{AC}, \vec{AD} \)
Given the points:
- \( A(0, 1, 2) \)
- \( B(-1, 2, 1) \)
- \( C(1, 2, 1) \)
- Let \( D \) have coordinates \( (x, y, z) \).

We calculate the vectors:
\[ \vec{AB} = B - A = (-1 - 0, 2 - 1, 1 - 2) = (-1, 1, -1) \]
\[ \vec{AC} = C - A = (1 - 0, 2 - 1, 1 - 2) = (1, 1, -1) \]
\[ \vec{AD} = D - A = (x - 0, y - 1, z - 2) = (x, y - 1, z - 2) \]

### Step 3: Calculate the Cross Product \( \vec{AC} \times \vec{AD} \)
Using the determinant to find the cross product:
\[ \vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ x & y - 1 & z - 2 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \left( 1(z - 2) - (-1)(y - 1) \right) - \hat{j} \left( 1(z - 2) - (-1)x \right) + \hat{k} \left( 1(y - 1) - 1x \right) \]
\[ = \hat{i} (z - 2 + y - 1) - \hat{j} (z - 2 + x) + \hat{k} (y - 1 - x) \]
\[ = \hat{i} (y + z - 3) - \hat{j} (z + x - 2) + \hat{k} (y - x - 1) \]

### Step 4: Calculate the Scalar Product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \)
Now we compute the scalar product:
\[ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (-1, 1, -1) \cdot \left( (y + z - 3), -(z + x - 2), (y - x - 1) \right) \]
Calculating this gives:
\[ = -1(y + z - 3) + 1(z + x - 2) - 1(y - x - 1) \]
\[ = -y - z + 3 + z + x - 2 - y + x + 1 \]
\[ = 2x - 2y + 2 \]

### Step 5: Set the Volume Equal to 1
Since the volume is given as 1 cubic unit:
\[ \frac{1}{6} |2x - 2y + 2| = 1 \]
This implies:
\[ |2x - 2y + 2| = 6 \]
This leads to two equations:
1. \( 2x - 2y + 2 = 6 \)
2. \( 2x - 2y + 2 = -6 \)

### Step 6: Solve the Equations
1. From \( 2x - 2y + 2 = 6 \):
\[ 2x - 2y = 4 \implies x - y = 2 \implies y = x - 2 \]

2. From \( 2x - 2y + 2 = -6 \):
\[ 2x - 2y = -8 \implies x - y = -4 \implies y = x + 4 \]

### Step 7: Find the Locus of Point \( D \)
Combining both equations, we can express the locus of point \( D \):
\[ y + z = 0 \quad \text{(from the first equation)} \]
Thus, the locus of point \( D \) is given by:
\[ y + z = 0 \]

### Final Answer
The locus of point \( D \) is:
**Option c: \( y + z = 0 \)**