For what value (s) of a will the two points `(1,a ,1)a n d(-3,0,a)` lie on opposite sides of the plane `3x+4y-12 z+13=0?` a. `a >-1ora >1//3` b. `a=0` only c. `0

Options

A. `alt-1oragt1//3`
B. a=0 only
C. `0ltalt1`
D. `-1ltalt1`

Explanation

To determine the values of \( a \) for which the points \( (1, a, 1) \) and \( (-3, 0, a) \) lie on opposite sides of the plane given by the equation \( 3x + 4y - 12z + 13 = 0 \), we will follow these steps:

### Step 1: Substitute the first point into the plane equation.
The first point is \( (1, a, 1) \). We substitute \( x = 1 \), \( y = a \), and \( z = 1 \) into the plane equation:

\[ 3(1) + 4(a) - 12(1) + 13 = 0 \]

This simplifies to:

\[ 3 + 4a - 12 + 13 = 0 \]

Combining like terms gives:

\[ 4a + 4 = 0 \]

### Step 2: Solve for \( a \) from the first point.
Rearranging the equation from Step 1:

\[ 4a = -4 \implies a = -1 \]

### Step 3: Substitute the second point into the plane equation.
Now we substitute the second point \( (-3, 0, a) \) into the plane equation:

\[ 3(-3) + 4(0) - 12(a) + 13 = 0 \]

This simplifies to:

\[ -9 - 12a + 13 = 0 \]

Combining like terms gives:

\[ -12a + 4 = 0 \]

### Step 4: Solve for \( a \) from the second point.
Rearranging the equation from Step 3:

\[ -12a = -4 \implies a = \frac{1}{3} \]

### Step 5: Determine the conditions for opposite sides.
For the points to lie on opposite sides of the plane, the results from substituting both points must yield opposite signs. We have:

1. For \( (1, a, 1) \): \( 4a + 4 \)
2. For \( (-3, 0, a) \): \( -12a + 4 \)

We need to analyze the signs of these expressions.

### Step 6: Set up inequalities.
We want:

\[ (4a + 4)(-12a + 4) < 0 \]

### Step 7: Find the critical points.
The critical points from the equations are:

1. \( 4a + 4 = 0 \) gives \( a = -1 \)
2. \( -12a + 4 = 0 \) gives \( a = \frac{1}{3} \)

### Step 8: Test intervals.
We will test the intervals defined by the critical points \( (-\infty, -1) \), \( (-1, \frac{1}{3}) \), and \( (\frac{1}{3}, \infty) \).

1. **Interval \( (-\infty, -1) \)**: Choose \( a = -2 \):
- \( 4(-2) + 4 = -8 + 4 = -4 \) (negative)
- \( -12(-2) + 4 = 24 + 4 = 28 \) (positive)
- Product: negative.

2. **Interval \( (-1, \frac{1}{3}) \)**: Choose \( a = 0 \):
- \( 4(0) + 4 = 4 \) (positive)
- \( -12(0) + 4 = 4 \) (positive)
- Product: positive.

3. **Interval \( (\frac{1}{3}, \infty) \)**: Choose \( a = 1 \):
- \( 4(1) + 4 = 8 \) (positive)
- \( -12(1) + 4 = -8 \) (negative)
- Product: negative.

### Conclusion:
The points \( (1, a, 1) \) and \( (-3, 0, a) \) lie on opposite sides of the plane when \( a < -1 \) or \( a > \frac{1}{3} \).

Thus, the solution is:

\[ a < -1 \text{ or } a > \frac{1}{3} \]

### Final Answer:
The correct option is:
**a. \( a > -1 \) or \( a > \frac{1}{3} \)**