The line passing through the point -1, 2, 3 and perpendicular to the plane x - 2y + 3z + 5 = 0 will be

Options

A. `(x + 1)/(1) = (y - 2)/(3) = (z - 3)/(5)`
B. `(x + 1)/(1) = (y - 2)/(3) = -(z - 3)/(3)`
C. `(x + 1)/(1) = (y - 2)/(3) = (z - 3)/(z)`
D. `(x + 1)/(1) = (y - 2)/(-2) = (z - 3)/(3)`

Answer

To find the equation of the line passing through the point (-1, 2, 3) and perpendicular to the plane given by the equation \(x - 2y + 3z + 5 = 0\), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The normal vector of a plane given by the equation \(Ax + By + Cz + D = 0\) can be determined from the coefficients of \(x\), \(y\), and \(z\). For the plane \(x - 2y + 3z + 5 = 0\), the coefficients are:
- \(A = 1\)
- \(B = -2\)
- \(C = 3\)

Thus, the normal vector \(\vec{n}\) of the plane is:
\[ \vec{n} = (1, -2, 3) \]

### Step 2: Write the parametric equations of the line
The line that is perpendicular to the plane will have the direction of the normal vector. The parametric equations of the line can be expressed as:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \]
where \((x_1, y_1, z_1)\) is the point through which the line passes, and \((a, b, c)\) are the components of the direction vector.

Substituting the point \((-1, 2, 3)\) and the normal vector \((1, -2, 3)\):
\[ \frac{x + 1}{1} = \frac{y - 2}{-2} = \frac{z - 3}{3} \]

### Step 3: Write the equations in a standard form
From the above, we can separate the equations:
1. \(x + 1 = t\) (where \(t\) is a parameter)
2. \(y - 2 = -2t\)
3. \(z - 3 = 3t\)

Rearranging these gives:
1. \(x = t - 1\)
2. \(y = -2t + 2\)
3. \(z = 3t + 3\)

### Step 4: Combine into a single equation
The equations can be combined into a single vector equation of the line:
\[ \vec{r} = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \]
This represents the line in vector form.

### Final Answer
The line passing through the point (-1, 2, 3) and perpendicular to the plane \(x - 2y + 3z + 5 = 0\) is given by:
\[ \vec{r} = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \]