Let `a_1x+b_1y+c_1z+d_1=0 and a_2x+b_2y+c_2z+d_2=0 ` be two planes, where `d_1, d_2gt0`. Then, origin lies in acute angle, If `a_1a_2+b_1b_2+c_1c_2lt0` and origin lies in obtuse angle if `a_1a_2+b_1b_2+c_1c_2gt0`.
Further point `(x_1, y_1, z_1)` and origin both lie either in acute angle or in obtuse angle. If ( `a_1x_1+b_1y_1+c_1z_1+d_1)(a_2x_1+b_2y_1+c_2z_1+d_2)gt0`.
one of `(x_1, y_1, z_1)` and origin in lie in acute and the other in obtuse angle,If ( `a_1x_1+b_1y_1+c_1z_1+d_1)(a_2x_1+b_2y_1+c_2z_1+d_2)lt0`
Q. Given that planes `2x+3y-4z+7=0 and x-2y+3z-5=0`. If a point `P(1, -2, 3),` then

Options

A. O and P both lie in acute angle between the planes
B. O and P both lies in obtuse angle
C. O lies in acute angle, P lies in obtuse angle
D. O lies in obtuse angle, P lies in acute angle

Answer

To solve the problem, we need to analyze the given planes and the point in relation to the conditions provided.

### Step-by-Step Solution:

1. **Identify the equations of the planes:**
- The first plane is given by:
\[ 2x + 3y - 4z + 7 = 0 \]
- The second plane is given by:
\[ x - 2y + 3z - 5 = 0 \]

2. **Check the conditions for \(d_1\) and \(d_2\):**
- From the first plane, \(d_1 = 7\) (which is greater than 0).
- From the second plane, \(d_2 = -5\). To make this positive, we can rewrite the equation as:
\[ -x + 2y - 3z + 5 = 0 \quad \text{(Multiplying the entire equation by -1)} \]
Here, \(d_2 = 5\) (which is also greater than 0).

3. **Calculate the dot product \(a_1a_2 + b_1b_2 + c_1c_2\):**
- For the first plane, \(a_1 = 2\), \(b_1 = 3\), \(c_1 = -4\).
- For the second plane, \(a_2 = 1\), \(b_2 = -2\), \(c_2 = 3\).
- Now calculate:
\[ a_1a_2 + b_1b_2 + c_1c_2 = (2)(1) + (3)(-2) + (-4)(3) \]
\[ = 2 - 6 - 12 = -16 \]
- Since \(-16 < 0\), the origin lies in the acute angle between the two planes.

4. **Evaluate the point \(P(1, -2, 3)\):**
- Substitute the coordinates of point \(P\) into the plane equations to check the signs:
- For the first plane:
\[ 2(1) + 3(-2) - 4(3) + 7 = 2 - 6 - 12 + 7 = -9 \]
- For the second plane:
\[ 1 - 2(-2) + 3(3) - 5 = 1 + 4 + 9 - 5 = 9 \]
- The results are:
- For the first plane: \(-9 < 0\)
- For the second plane: \(9 > 0\)

5. **Check the product of the results:**
- Now we check the product of the results:
\[ (-9)(9) = -81 < 0 \]
- This indicates that one of the points (the origin) lies in the acute angle and the other point \(P\) lies in the obtuse angle.

### Conclusion:
- The origin lies in the acute angle while point \(P\) lies in the obtuse angle, confirming the conditions given in the problem.

### Final Answer:
- The answer is that the origin lies in the acute angle and point \(P(1, -2, 3)\) lies in the obtuse angle.

---