Suppose `(alpha,beta,gamma)` lie on the plane 2x+y+z=1 and `[alpha,beta,gamma][(1,9,1),(7,2,1),(8,3,1)]=[0,0,0]` then `alpha +beta^(2)+gamma^(2)` = _____

Answer

To solve the problem step by step, we need to find the values of \( \alpha \), \( \beta \), and \( \gamma \) given the conditions and then calculate \( \alpha + \beta^2 + \gamma^2 \).

### Step 1: Understand the plane equation
We know that \( \alpha, \beta, \gamma \) lie on the plane defined by the equation:
\[ 2\alpha + \beta + \gamma = 1 \]

### Step 2: Set up the matrix equation
We also have the matrix equation:
\[ [\alpha, \beta, \gamma] \begin{pmatrix} 1 & 9 & 1 \\ 7 & 2 & 1 \\ 8 & 3 & 1 \end{pmatrix} = [0, 0, 0] \]
This expands to three equations:
1. \( \alpha + 9\beta + \gamma = 0 \) (Equation 1)
2. \( 7\alpha + 2\beta + \gamma = 0 \) (Equation 2)
3. \( 8\alpha + 3\beta + \gamma = 0 \) (Equation 3)

### Step 3: Solve the equations
From Equation 1, we can express \( \gamma \) in terms of \( \alpha \) and \( \beta \):
\[ \gamma = -\alpha - 9\beta \]

Substituting this into Equation 2:
\[ 7\alpha + 2\beta + (-\alpha - 9\beta) = 0 \]
This simplifies to:
\[ 6\alpha - 7\beta = 0 \implies 6\alpha = 7\beta \implies \beta = \frac{6}{7}\alpha \]

Now substitute \( \beta \) back into the expression for \( \gamma \):
\[ \gamma = -\alpha - 9\left(\frac{6}{7}\alpha\right) = -\alpha - \frac{54}{7}\alpha = -\frac{61}{7}\alpha \]

### Step 4: Substitute into the plane equation
Now substitute \( \beta \) and \( \gamma \) into the plane equation:
\[ 2\alpha + \frac{6}{7}\alpha - \frac{61}{7}\alpha = 1 \]
Combining the terms:
\[ 2\alpha + \frac{6\alpha - 61\alpha}{7} = 1 \implies 2\alpha - \frac{55}{7}\alpha = 1 \]
To eliminate the fraction, multiply through by 7:
\[ 14\alpha - 55\alpha = 7 \implies -41\alpha = 7 \implies \alpha = -\frac{7}{41} \]

### Step 5: Find \( \beta \) and \( \gamma \)
Now substitute \( \alpha \) back to find \( \beta \) and \( \gamma \):
\[ \beta = \frac{6}{7}\left(-\frac{7}{41}\right) = -\frac{6}{41} \]
\[ \gamma = -\frac{61}{7}\left(-\frac{7}{41}\right) = \frac{61}{41} \]

### Step 6: Calculate \( \alpha + \beta^2 + \gamma^2 \)
Now we calculate \( \alpha + \beta^2 + \gamma^2 \):
\[ \beta^2 = \left(-\frac{6}{41}\right)^2 = \frac{36}{1681} \]
\[ \gamma^2 = \left(\frac{61}{41}\right)^2 = \frac{3721}{1681} \]
Now, adding these:
\[ \alpha + \beta^2 + \gamma^2 = -\frac{7}{41} + \frac{36}{1681} + \frac{3721}{1681} \]
Convert \( -\frac{7}{41} \) to a fraction with a denominator of 1681:
\[ -\frac{7}{41} = -\frac{7 \times 41}{41 \times 41} = -\frac{287}{1681} \]
Now combine:
\[ -\frac{287}{1681} + \frac{36 + 3721}{1681} = -\frac{287}{1681} + \frac{3757}{1681} = \frac{3757 - 287}{1681} = \frac{3470}{1681} \]

### Final Answer
Thus, \( \alpha + \beta^2 + \gamma^2 = \frac{3470}{1681} \).