If the straight line `(x-1)/3=(y-2)/2=(z-3)/2` intersect the curve `ax^(2) + by^(2) = 1, z = 0` find the value of `49a + 4b.`

Answer

To solve the problem, we need to find the intersection of the given straight line with the curve defined by the equation \( ax^2 + by^2 = 1 \) at the plane \( z = 0 \).

### Step-by-Step Solution:

1. **Parameterize the Straight Line**:
The given line is represented as:
\[ \frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{2} \]
We can introduce a parameter \( \lambda \) such that:
\[ x = 3\lambda + 1, \quad y = 2\lambda + 2, \quad z = 2\lambda + 3 \]

2. **Set the z-coordinate to 0**:
Since we are interested in the intersection with the plane \( z = 0 \), we set:
\[ 2\lambda + 3 = 0 \]
Solving for \( \lambda \):
\[ 2\lambda = -3 \implies \lambda = -\frac{3}{2} \]

3. **Substitute \( \lambda \) back to find x and y**:
Now, substitute \( \lambda = -\frac{3}{2} \) into the equations for \( x \) and \( y \):
\[ x = 3\left(-\frac{3}{2}\right) + 1 = -\frac{9}{2} + 1 = -\frac{9}{2} + \frac{2}{2} = -\frac{7}{2} \]
\[ y = 2\left(-\frac{3}{2}\right) + 2 = -3 + 2 = -1 \]
Thus, the point of intersection \( Q \) is:
\[ Q\left(-\frac{7}{2}, -1, 0\right) \]

4. **Substitute into the curve equation**:
We substitute the coordinates of point \( Q \) into the curve equation \( ax^2 + by^2 = 1 \):
\[ a\left(-\frac{7}{2}\right)^2 + b(-1)^2 = 1 \]
This simplifies to:
\[ a\left(\frac{49}{4}\right) + b(1) = 1 \]
or:
\[ \frac{49a}{4} + b = 1 \]

5. **Multiply through by 4 to eliminate the fraction**:
Multiply the entire equation by 4:
\[ 49a + 4b = 4 \]

6. **Final Result**:
We need the value of \( 49a + 4b \):
\[ 49a + 4b = 4 \]

### Conclusion:
The value of \( 49a + 4b \) is \( 4 \).