The equation of the line through the point (0,1,2) and perpendicular to the line `(x-1)/(2)=(y+1)/(3)=(z-1)/(-2)` is :

Options

A. `(x)/(3)=(y-1)/(4)=(z-2)/(3)`
B. `(x)/(3)=(y-1)/(-4)=(z-2)/(3)`
C. `(x)/(3)=(y-1)/(4)=(z-2)/(-3)`
D. `(x)/(-3)=(y-1)/(4)=(z-2)/(3)`

Answer

To find the equation of the line that passes through the point (0, 1, 2) and is perpendicular to the given line, we can follow these steps:

### Step 1: Identify the direction vector of the given line
The given line is represented in symmetric form as:
\[ \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2} \]
From this, we can extract the direction vector of the line, which is:
\[ \mathbf{a_1} = \langle 2, 3, -2 \rangle \]

### Step 2: Set up the direction vector of the new line
Let the direction vector of the new line be \(\mathbf{b_1} = \langle a, b, c \rangle\). Since the new line is perpendicular to the given line, the dot product of the two direction vectors must equal zero:
\[ \mathbf{a_1} \cdot \mathbf{b_1} = 0 \]
This gives us the equation:
\[ 2a + 3b - 2c = 0 \]

### Step 3: Choose a direction vector that satisfies the perpendicular condition
We can choose specific values for \(a\), \(b\), and \(c\) to satisfy the equation. For example, let’s set \(a = 3\) and \(b = -2\). Then we can solve for \(c\):
\[ 2(3) + 3(-2) - 2c = 0 \implies 6 - 6 - 2c = 0 \implies -2c = 0 \implies c = 0 \]
Thus, one possible direction vector for the new line is:
\[ \mathbf{b_1} = \langle 3, -2, 0 \rangle \]

### Step 4: Write the equation of the new line
The equation of a line in vector form that passes through a point \((x_0, y_0, z_0)\) with direction vector \(\langle a, b, c \rangle\) is given by:
\[ \mathbf{r} = \mathbf{r_0} + t \mathbf{b_1} \]
where \(\mathbf{r_0} = \langle 0, 1, 2 \rangle\) and \(\mathbf{b_1} = \langle 3, -2, 0 \rangle\). Thus, the equation of the line is:
\[ \mathbf{r} = \langle 0, 1, 2 \rangle + t \langle 3, -2, 0 \rangle \]
This can be expressed in parametric form as:
\[ x = 3t, \quad y = 1 - 2t, \quad z = 2 \]

### Final Answer
The equation of the line through the point (0, 1, 2) and perpendicular to the line is:
\[ \begin{cases} x = 3t \\ y = 1 - 2t \\ z = 2 \end{cases} \]

---