Find the Cartesian equation of the plane, passing through the line of intersection of the planes:
`vecr.(2hati+3hatj-4hatk)+5=0 and vecr.(hati-5hatj+7hatk)+2=0` and intersecting Y-axis at (0, 3,0).

Answer

To find the Cartesian equation of the plane passing through the line of intersection of the given planes and intersecting the Y-axis at the point (0, 3, 0), we can follow these steps:

### Step 1: Write the equations of the given planes
The equations of the given planes are:
1. \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) + 5 = 0 \)
2. \( \vec{r} \cdot (\hat{i} - 5\hat{j} + 7\hat{k}) + 2 = 0 \)

### Step 2: General equation of the plane
The general equation of the plane that passes through the line of intersection of the two planes can be expressed as:
\[ 2x + 3y - 4z + 5 + \lambda (x - 5y + 7z + 2) = 0 \]
where \( \lambda \) is a parameter.

### Step 3: Simplify the equation
Expanding the equation:
\[ 2x + 3y - 4z + 5 + \lambda x - 5\lambda y + 7\lambda z + 2\lambda = 0 \]
Combining like terms gives:
\[ (2 + \lambda)x + (3 - 5\lambda)y + (-4 + 7\lambda)z + (5 + 2\lambda) = 0 \]

### Step 4: Substitute the point (0, 3, 0)
Since the plane intersects the Y-axis at the point (0, 3, 0), we substitute \( x = 0 \), \( y = 3 \), and \( z = 0 \) into the equation:
\[ (2 + \lambda)(0) + (3 - 5\lambda)(3) + (-4 + 7\lambda)(0) + (5 + 2\lambda) = 0 \]
This simplifies to:
\[ (3 - 5\lambda)(3) + (5 + 2\lambda) = 0 \]
\[ 9 - 15\lambda + 5 + 2\lambda = 0 \]
\[ 14 - 13\lambda = 0 \]

### Step 5: Solve for \( \lambda \)
From the equation \( 14 - 13\lambda = 0 \), we solve for \( \lambda \):
\[ 13\lambda = 14 \implies \lambda = \frac{14}{13} \]

### Step 6: Substitute \( \lambda \) back into the plane equation
Now, substitute \( \lambda = \frac{14}{13} \) back into the general equation:
\[ (2 + \frac{14}{13})x + (3 - 5 \cdot \frac{14}{13})y + (-4 + 7 \cdot \frac{14}{13})z + (5 + 2 \cdot \frac{14}{13}) = 0 \]
Calculating each coefficient:
- For \( x \): \( 2 + \frac{14}{13} = \frac{26}{13} + \frac{14}{13} = \frac{40}{13} \)
- For \( y \): \( 3 - \frac{70}{13} = \frac{39}{13} - \frac{70}{13} = -\frac{31}{13} \)
- For \( z \): \( -4 + \frac{98}{13} = -\frac{52}{13} + \frac{98}{13} = \frac{46}{13} \)
- Constant term: \( 5 + \frac{28}{13} = \frac{65}{13} + \frac{28}{13} = \frac{93}{13} \)

### Step 7: Final equation of the plane
Thus, the equation of the plane becomes:
\[ \frac{40}{13}x - \frac{31}{13}y + \frac{46}{13}z + \frac{93}{13} = 0 \]
Multiplying through by 13 to eliminate the fraction:
\[ 40x - 31y + 46z + 93 = 0 \]

### Final Answer
The Cartesian equation of the plane is:
\[ 40x - 31y + 46z + 93 = 0 \]