Answer :

given pair of linear equations are

(p – 3) x + 3y = p

⇒ (p – 3) x + 3y – p = 0

px + py = 12

⇒ px + py – 12 = 0

Comparing these with standard equation

a_{1}x_{1} + b_{1}x_{1} + c_{1} = 0 and a_{2}x_{2} + b_{2}x_{2} + c_{2} = 0

We get

a_{1} = (p – 3), b_{1} = 3, c_{1} = – p

a_{2} = p, b_{2} = p, c_{2} = – 12

Now it is given the linear pair of equations have infinitely many solutions, so the condition for it is:

Now substituting the corresponding values, we get

Considering first equality, we get

⇒ p – 3 = 3

⇒ p = 6

Considering the second equality, we get

⇒ 3 × 12 = p^{2}

⇒ p = √36

⇒ p = ±6

Hence p = + 6 satisfies both the conditions.

**Hence for p = 6, the given pair of linear equations have infinitely many solutions.**

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