If the three planes `x=5,2x-5ay+3z-2=0` and `3bx+y-3z=0` contain a common line, then `(a,b)` is equal to

Options

A. `(-1/5,8/15)`
B. `(1/5,-8/15)`
C. `(-8/15,1/5)`
D. `(8/15,-1/5)`

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) such that the three given planes contain a common line. The planes are given as:

1. \( x = 5 \)
2. \( 2x - 5ay + 3z - 2 = 0 \)
3. \( 3bx + y - 3z = 0 \)

### Step 1: Substitute \( x = 5 \) into the equations of the planes.

For the second plane:
\[ 2(5) - 5ay + 3z - 2 = 0 \]
This simplifies to:
\[ 10 - 5ay + 3z - 2 = 0 \implies 8 - 5ay + 3z = 0 \tag{1} \]

For the third plane:
\[ 3b(5) + y - 3z = 0 \]
This simplifies to:
\[ 15b + y - 3z = 0 \tag{2} \]

### Step 2: Rearranging the equations.

From equation (1):
\[ -5ay + 3z = -8 \implies 5ay - 3z = 8 \tag{3} \]

From equation (2):
\[ y - 3z = -15b \implies y - 3z + 15b = 0 \tag{4} \]

### Step 3: Compare coefficients.

Since both equations (3) and (4) represent the same line, we can compare the coefficients of \( y \) and \( z \).

From equation (3):
- Coefficient of \( y \) is \( 5a \)
- Coefficient of \( z \) is \( -3 \)

From equation (4):
- Coefficient of \( y \) is \( 1 \)
- Coefficient of \( z \) is \( -3 \)

Setting the coefficients equal gives us:
\[ 5a = 1 \implies a = \frac{1}{5} \]

### Step 4: Compare the constant terms.

Now, we can compare the constant terms from equations (3) and (4):
From equation (3), the constant term is \( 8 \).
From equation (4), the constant term is \( -15b \).

Setting these equal gives us:
\[ 8 = -15b \implies b = -\frac{8}{15} \]

### Conclusion

Thus, the values of \( (a, b) \) are:
\[ (a, b) = \left( \frac{1}{5}, -\frac{8}{15} \right) \]