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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