Step 1: Form equations
Let orange \(=O\), banana \(=B\), apple \(=A\).
Equation (1): \(2O+3B+4A=15\).
Equation (2): \(3O+2B+A=10\).
Step 2: Required expression
We need \(3O+3B+3A\).
Step 3: Manipulate equations
Multiply (2) by 3: \(9O+6B+3A=30\).
Multiply (1) by 3: \(6O+9B+12A=45\).
Subtract second from first: \((9O-6O)+(6B-9B)+(3A-12A)=30-45\).
\(⇒ 3O-3B-9A=-15\ \(⇒\) O-B-3A=-5\).
This single relation plus the originals allows solving for \(3O+3B+3A\). Add Eqn (1) and Eqn (2): \((2O+3O)+(3B+2B)+(4A+A)=15+10\).
\(⇒ 5O+5B+5A=25 \(⇒\) O+B+A=5\).
So \(3O+3B+3A=15\).
\[
\boxed{₹ 15}
\]