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)

Explanation

To find where the line meets the plane given the point (1, -2, 3) and the plane equation \(2x + 3y - z = 7\), we can follow these steps:

### Step 1: Identify the direction ratios of the line
Since the line is perpendicular to the plane, its direction ratios are the same as the coefficients of \(x\), \(y\), and \(z\) in the plane equation. The plane equation is \(2x + 3y - z = 7\), so the direction ratios of the line are:
- \(a = 2\)
- \(b = 3\)
- \(c = -1\)

### Step 2: Write the parametric equations of the line
Using the point (1, -2, 3) and the direction ratios, we can write the parametric equations of the line:
\[ \frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{-1} = t \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(t\):
- \(x = 1 + 2t\)
- \(y = -2 + 3t\)
- \(z = 3 - t\)

### Step 3: Substitute the parametric equations into the plane equation
Now, we need 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
Now, simplify the equation:
\[ 2 + 4t - 6 + 9t - 3 + t = 7 \]
Combine like terms:
\[ (4t + 9t + t) + (2 - 6 - 3) = 7 \]
\[ 14t - 7 = 7 \]

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

### Step 6: Find the coordinates of the intersection point
Now, substitute \(t = 1\) 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 line meets the plane at the point \((3, 1, 2)\).

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

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