For some constant real numbers p,k and a ,consider the following system of linear equations in x and y : px−4y=2, 3x+ky=a A necessary condition for the system to have no solution for (x,y) , is

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The given system of linear equations is:

$$
px - 4y = 2 \quad 	ext{(Equation 1)}
$$
$$
3x + ky = a \quad 	ext{(Equation 2)}
$$

For the system to have no solution, the lines represented by these equations must be parallel but not coincident. This occurs when:

$$
\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}
$$

Here, $$a_1, b_1, c_1$$ are the coefficients of $$x, y,$$ and the constant term in Equation 1, respectively, and $$a_2, b_2, c_2$$ are the corresponding coefficients in Equation 2.

From the equations:
- $$a_1 = p, b_1 = -4, c_1 = -2$$
- $$a_2 = 3, b_2 = k, c_2 = -a$$

Substituting into the condition:

$$
\frac{p}{3} = \frac{-4}{k} 
eq \frac{-2}{-a}
$$

Breaking this into cases:

1. **Case 1: $$\frac{p}{3} = \frac{-4}{k}$$:**
   Solving for $$k$$,
   $$
   pk + 12 = 0
   $$

2. **Case 2: $$\frac{-4}{k} 
eq \frac{-2}{-a}$$:**
   Simplifying,
   $$
   -4 
eq 2k/a \quad \Rightarrow \quad 2a + k 
eq 0
   $$

3. **Case 3: $$\frac{p}{3} 
eq \frac{-2}{-a}$$:**
   Simplifying,
   $$
   pa - 6 
eq 0
   $$

Among these conditions, the necessary condition for no solution is:

$$
2a + k 
eq 0
$$

Thus, **Option B** is correct.

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