1. **Problem Statement:** Find the new optimal solution for the linear program when the objective function changes to: (a) $z = 12 x_1 + 5 x_2 + 2 x_3$ 2. **Given:** Original optimal tableau: \begin{array}{c|cccccc} \text{Basis} & x_1 & x_2 & x_3 & R & x_4 & \text{Right Side} \\ \hline Z & 0 & 23 & 7 & 5 + M & 0 & 150 \\ x_1 & 1 & 5 & 2 & 1 & 0 & 30 \\ x_2 & 0 & -10 & -8 & -1 & 1 & 10 \end{array} 3. **Step 1: Understand the problem** We want to find the new optimal value of $z$ when the objective function coefficients change to $12, 5, 2$ for $x_1, x_2, x_3$ respectively. 4. **Step 2: Use sensitivity analysis formula** The new objective function coefficients vector is $c' = (12, 5, 2)$. The current basic variables are $x_1$ and $x_2$ with values 30 and 10 respectively. The new optimal value can be found by: $$ z' = c'_B \cdot x_B $$ where $c'_B$ are the new coefficients of basic variables and $x_B$ are their values. 5. **Step 3: Identify basic variables and their new coefficients** Basic variables: $x_1$ and $x_2$ New coefficients: $c'_{x_1} = 12$, $c'_{x_2} = 5$ 6. **Step 4: Calculate new optimal value** $$ z' = 12 \times 30 + 5 \times 10 = 360 + 50 = 410 $$ 7. **Answer:** The new optimal value of $z$ is **410** when $z = 12 x_1 + 5 x_2 + 2 x_3$. --- **Note:** The other parts of the question are not solved as per instructions to solve only the first problem.