A plane passing through the point (3,1,1) contains two lines whose direction ratios are 1,–2,2 and 2,3, –1 respectively. If this plane also passes through the point `(alpha, –3,5)` , then `alpha` is equal to:

Options

A. -5
B. 10
C. 5
D. -10

Answer

To solve the problem step by step, we need to find the value of \( \alpha \) such that the point \( (\alpha, -3, 5) \) lies on the plane defined by the given conditions.

### Step 1: Identify the plane equation
The plane passes through the point \( (3, 1, 1) \) and contains two lines with direction ratios \( (1, -2, 2) \) and \( (2, 3, -1) \). The normal vector to the plane can be found by taking the cross product of the direction ratios of the two lines.

Let:
- \( \mathbf{d_1} = (1, -2, 2) \)
- \( \mathbf{d_2} = (2, 3, -1) \)

### Step 2: Calculate the cross product
The cross product \( \mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} \) gives us the normal vector of the plane.

\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 2 \\ 2 & 3 & -1 \end{vmatrix} \]

Calculating this determinant:

\[ \mathbf{n} = \mathbf{i}((-2)(-1) - (2)(3)) - \mathbf{j}((1)(-1) - (2)(2)) + \mathbf{k}((1)(3) - (-2)(2)) \]
\[ = \mathbf{i}(2 - 6) - \mathbf{j}(-1 - 4) + \mathbf{k}(3 + 4) \]
\[ = \mathbf{i}(-4) + \mathbf{j}(5) + \mathbf{k}(7) \]

Thus, the normal vector \( \mathbf{n} = (-4, 5, 7) \).

### Step 3: Write the equation of the plane
Using the point-normal form of the plane equation, we have:

\[ -4(x - 3) + 5(y - 1) + 7(z - 1) = 0 \]

Expanding this:

\[ -4x + 12 + 5y - 5 + 7z - 7 = 0 \]
\[ -4x + 5y + 7z = 0 \]

### Step 4: Substitute the point \( (\alpha, -3, 5) \)
Now, we substitute the point \( (\alpha, -3, 5) \) into the plane equation:

\[ -4(\alpha) + 5(-3) + 7(5) = 0 \]

Calculating this:

\[ -4\alpha - 15 + 35 = 0 \]
\[ -4\alpha + 20 = 0 \]
\[ -4\alpha = -20 \]
\[ \alpha = 5 \]

### Conclusion
The value of \( \alpha \) is \( 5 \).