A straight line passes through `(1,-2,3)` and perpendicular to the plane `2x+3y-z=7`
Where does the line meet the plane ?

Options

A. `(2,3,1)`
B. `(1,2,3)`
C. `(2,1,3)`
D. `(3,1,2)`

Answer

To find where the line meets the plane given that it passes through the point (1, -2, 3) and is perpendicular to the plane defined by the equation \(2x + 3y - z = 7\), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The equation of the plane is given as \(2x + 3y - z = 7\). The coefficients of \(x\), \(y\), and \(z\) in this equation represent the components of the normal vector to the plane. Thus, the normal vector \( \mathbf{n} \) is:
\[ \mathbf{n} = (2, 3, -1) \]

### Step 2: Write the parametric equations of the line
Since the line is perpendicular to the plane, it will have the same direction as the normal vector. The parametric equations of the line passing through the point (1, -2, 3) can be expressed as:
\[ x = 1 + 2t \]
\[ y = -2 + 3t \]
\[ z = 3 - t \]
where \(t\) is a parameter.

### Step 3: Substitute the parametric equations into the plane equation
To find the point where the line intersects the plane, we substitute the parametric equations into the plane equation \(2x + 3y - z = 7\):
\[ 2(1 + 2t) + 3(-2 + 3t) - (3 - t) = 7 \]

### Step 4: Simplify the equation
Expanding the left side:
\[ 2 + 4t - 6 + 9t - 3 + t = 7 \]
Combining like terms:
\[ (4t + 9t + t) + (2 - 6 - 3) = 7 \]
\[ 14t - 7 = 7 \]

### Step 5: Solve for \(t\)
Now, we solve for \(t\):
\[ 14t = 7 + 7 \]
\[ 14t = 14 \]
\[ t = 1 \]

### Step 6: Find the coordinates of the intersection point
Now that we have \(t = 1\), we can substitute it back into the parametric equations to find the coordinates of the intersection point:
\[ x = 1 + 2(1) = 3 \]
\[ y = -2 + 3(1) = 1 \]
\[ z = 3 - (1) = 2 \]

Thus, the point where the line meets the plane is:
\[ (3, 1, 2) \]

### Final Answer
The line meets the plane at the point \((3, 1, 2)\).

---