Let `P=0` be the equation of a plane passing through the line of intersection of the planes `2x-y=0a n d3z-y=0` and perpendicular to the plane `4x+5y-3z=8.` Then the points which lie on the plane `P=0` is/are a. `(0,9,17)` b. `(1//7,2,1//9)` c. `(1,3,-4)` d. `(1//2,1,1//3)`

Options

A. (0,9,17)
B. `(1//7,2,1//9)`
C. (1,3,-4)
D. `(1//2,1,1//3)`

Explanation

To solve the problem, we need to find the equation of a plane that passes through the line of intersection of the planes given by \(2x - y = 0\) and \(3z - y = 0\), and is also perpendicular to the plane given by \(4x + 5y - 3z = 8\).

### Step 1: Find the equation of the plane through the line of intersection

The equation of a plane passing through the line of intersection of two planes can be expressed as:
\[ P = \lambda (2x - y) + \mu (3z - y) = 0 \]
where \(\lambda\) and \(\mu\) are scalars.

Expanding this, we have:
\[ P = 2\lambda x - \lambda y + 3\mu z - \mu y = 0 \]
This simplifies to:
\[ 2\lambda x + (-\lambda - \mu)y + 3\mu z = 0 \]

### Step 2: Find the normal vector of the given plane

The normal vector of the plane \(4x + 5y - 3z = 8\) is given by the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n_1} = (4, 5, -3) \]

### Step 3: Find the normal vector of the plane \(P\)

The normal vector of the plane \(P\) is:
\[ \mathbf{n_2} = (2\lambda, -\lambda - \mu, 3\mu) \]

### Step 4: Set up the perpendicularity condition

Since the plane \(P\) is perpendicular to the plane \(4x + 5y - 3z = 8\), the dot product of their normal vectors must be zero:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 0 \]
This gives us:
\[ (4, 5, -3) \cdot (2\lambda, -\lambda - \mu, 3\mu) = 0 \]
Calculating the dot product:
\[ 4(2\lambda) + 5(-\lambda - \mu) - 3(3\mu) = 0 \]
This simplifies to:
\[ 8\lambda - 5\lambda - 5\mu - 9\mu = 0 \]
\[ 3\lambda - 14\mu = 0 \]
From this, we can express \(\mu\) in terms of \(\lambda\):
\[ \mu = \frac{3}{14}\lambda \]

### Step 5: Substitute \(\mu\) back into the plane equation

Substituting \(\mu\) back into the equation of the plane:
\[ P = 2\lambda x - \lambda y + 3\left(\frac{3}{14}\lambda\right)z - \frac{3}{14}\lambda y = 0 \]
This simplifies to:
\[ 2\lambda x - \lambda y - \frac{3}{14}\lambda y + \frac{9}{14}\lambda z = 0 \]
Factoring out \(\lambda\):
\[ \lambda(2x - y - \frac{3}{14}y + \frac{9}{14}z) = 0 \]
This leads to:
\[ 2x - \frac{17}{14}y + \frac{9}{14}z = 0 \]
Multiplying through by 14 to eliminate the fractions:
\[ 28x - 17y + 9z = 0 \]

### Step 6: Check which points satisfy the plane equation

Now we will check each given point to see if it satisfies the equation \(28x - 17y + 9z = 0\).

1. **Point A: \((0, 9, 17)\)**
\[ 28(0) - 17(9) + 9(17) = 0 - 153 + 153 = 0 \quad \text{(Satisfies)} \]

2. **Point B: \(\left(\frac{1}{7}, 2, \frac{1}{9}\right)\)**
\[ 28\left(\frac{1}{7}\right) - 17(2) + 9\left(\frac{1}{9}\right) = 4 - 34 + 1 = -29 \quad \text{(Does not satisfy)} \]

3. **Point C: \((1, 3, -4)\)**
\[ 28(1) - 17(3) + 9(-4) = 28 - 51 - 36 = -59 \quad \text{(Does not satisfy)} \]

4. **Point D: \(\left(\frac{1}{2}, 1, \frac{1}{3}\right)\)**
\[ 28\left(\frac{1}{2}\right) - 17(1) + 9\left(\frac{1}{3}\right) = 14 - 17 + 3 = 0 \quad \text{(Satisfies)} \]

### Conclusion

The points that lie on the plane \(P = 0\) are:
- Point A: \((0, 9, 17)\)
- Point D: \(\left(\frac{1}{2}, 1, \frac{1}{3}\right)\)