Consider a plane `P:2x+y-z=5`, a line `L:(x-3)/(2)=(y+1)/(-3)=(z-2)/(-1)` and a point `A(3, -4,1)`. If the line L intersects plane P at B and the xy plane at C, then the area (in sq. units) of `DeltaABC` is

Options

A. `sqrt7`
B. `sqrt8`
C. `sqrt(10)`
D. `2sqrt3`

Answer

To solve the problem, we need to find the area of triangle ABC formed by the points A, B, and C, where:

- A is given as \( A(3, -4, 1) \).
- B is the intersection of the line \( L \) with the plane \( P \).
- C is the intersection of the line \( L \) with the xy-plane.

### Step 1: Find the intersection point C of line L with the xy-plane

The equation of the line \( L \) is given as:
\[ \frac{x-3}{2} = \frac{y+1}{-3} = \frac{z-2}{-1} \]

To find the intersection with the xy-plane, we set \( z = 0 \).

From the line equation, we have:
\[ \frac{z-2}{-1} = \frac{0-2}{-1} = 2 \]
Thus, we can set:
\[ \frac{x-3}{2} = 2 \quad \text{and} \quad \frac{y+1}{-3} = 2 \]

Now, solving for \( x \):
\[ \frac{x-3}{2} = 2 \implies x - 3 = 4 \implies x = 7 \]

And solving for \( y \):
\[ \frac{y+1}{-3} = 2 \implies y + 1 = -6 \implies y = -7 \]

Thus, the coordinates of point \( C \) are:
\[ C(7, -7, 0) \]

### Step 2: Find the intersection point B of line L with plane P

The equation of the plane \( P \) is:
\[ 2x + y - z = 5 \]

We will substitute the parametric equations of the line \( L \) into the plane equation. The parametric equations from the line are:
\[ x = 2\lambda + 3, \quad y = -3\lambda - 1, \quad z = -\lambda + 2 \]

Substituting these into the plane equation:
\[ 2(2\lambda + 3) + (-3\lambda - 1) - (-\lambda + 2) = 5 \]

Expanding this:
\[ 4\lambda + 6 - 3\lambda - 1 + \lambda - 2 = 5 \]
Combining like terms:
\[ (4\lambda - 3\lambda + \lambda) + (6 - 1 - 2) = 5 \]
This simplifies to:
\[ 2\lambda + 3 = 5 \]
Solving for \( \lambda \):
\[ 2\lambda = 2 \implies \lambda = 1 \]

Now substituting \( \lambda = 1 \) back into the parametric equations to find \( B \):
\[ x = 2(1) + 3 = 5, \quad y = -3(1) - 1 = -4, \quad z = -1 + 2 = 1 \]

Thus, the coordinates of point \( B \) are:
\[ B(5, -4, 1) \]

### Step 3: Calculate the area of triangle ABC

We have:
- \( A(3, -4, 1) \)
- \( B(5, -4, 1) \)
- \( C(7, -7, 0) \)

Now, we can find vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):
\[ \overrightarrow{AB} = B - A = (5 - 3, -4 + 4, 1 - 1) = (2, 0, 0) \]
\[ \overrightarrow{AC} = C - A = (7 - 3, -7 + 4, 0 - 1) = (4, -3, -1) \]

Next, we will find the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \):
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 4 & -3 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(0 \cdot -1 - 0 \cdot -3) - \hat{j}(2 \cdot -1 - 0 \cdot 4) + \hat{k}(2 \cdot -3 - 0 \cdot 4) \]
\[ = \hat{i}(0) - \hat{j}(-2) + \hat{k}(-6) \]
\[ = 2\hat{j} - 6\hat{k} \]

The magnitude of the cross product is:
\[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{0^2 + 2^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \]

The area of triangle \( ABC \) is given by:
\[ \text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \cdot 2\sqrt{10} = \sqrt{10} \]

### Final Answer
The area of triangle \( ABC \) is:
\[ \boxed{\sqrt{10}} \text{ square units} \]