The linee joining the points (1,1,2) and (3,-2,1) meets the plane 3x+2y+z=6 at the point

Options

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

Answer

To find the point where the line joining the points \( A(1, 1, 2) \) and \( B(3, -2, 1) \) meets the plane given by the equation \( 3x + 2y + z = 6 \), we can follow these steps:

### Step 1: Determine the Direction Ratios of the Line
The direction ratios of the line joining points \( A(1, 1, 2) \) and \( B(3, -2, 1) \) can be found as follows:

\[ \text{Direction Ratios} = (3 - 1, -2 - 1, 1 - 2) = (2, -3, -1) \]

### Step 2: Write the Parametric Equations of the Line
Using point \( A(1, 1, 2) \) and the direction ratios, we can express the line in parametric form:

\[ \frac{x - 1}{2} = \frac{y - 1}{-3} = \frac{z - 2}{-1} = t \]

From this, we can derive the parametric equations:

\[ x = 2t + 1, \quad y = -3t + 1, \quad z = -t + 2 \]

### Step 3: Substitute the Parametric Equations into the Plane Equation
We substitute the parametric equations into the plane equation \( 3x + 2y + z = 6 \):

\[ 3(2t + 1) + 2(-3t + 1) + (-t + 2) = 6 \]

### Step 4: Simplify the Equation
Expanding and simplifying the equation:

\[ 6t + 3 - 6t + 2 - t + 2 = 6 \]
\[ 3 + 2 + 2 - t = 6 \]
\[ 7 - t = 6 \]

### Step 5: Solve for \( t \)
Rearranging gives:

\[ -t = 6 - 7 \]
\[ -t = -1 \implies t = 1 \]

### Step 6: Find the Point of Intersection
Now, substitute \( t = 1 \) back into the parametric equations to find the coordinates of the intersection point:

\[ x = 2(1) + 1 = 3 \]
\[ y = -3(1) + 1 = -2 \]
\[ z = -1 + 2 = 1 \]

Thus, the point of intersection is:

\[ (3, -2, 1) \]

### Final Answer
The line joining the points \( (1, 1, 2) \) and \( (3, -2, 1) \) meets the plane \( 3x + 2y + z = 6 \) at the point \( (3, -2, 1) \).

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