Let `vecA=hati+hat5j+alphahatk , vecB=hati+3hatj+betahatk , vecC=hati-2hatj+3hatk and abs(vecB xxvecC)=5sqrt3` . If `vecA` is perpendicular to `vecB` then the maxmimum value of `abs(vecA)^2` is

Answer

To solve the problem step by step, we need to follow the given conditions and perform calculations accordingly.

### Step 1: Define the vectors
We have three vectors defined as follows:
- \(\vec{A} = \hat{i} + 5\hat{j} + \alpha\hat{k}\)
- \(\vec{B} = \hat{i} + 3\hat{j} + \beta\hat{k}\)
- \(\vec{C} = \hat{i} - 2\hat{j} + 3\hat{k}\)

### Step 2: Use the condition of perpendicularity
Since \(\vec{A}\) is perpendicular to \(\vec{B}\), their dot product must equal zero:
\[ \vec{A} \cdot \vec{B} = 0 \]
Calculating the dot product:
\[ (\hat{i} + 5\hat{j} + \alpha\hat{k}) \cdot (\hat{i} + 3\hat{j} + \beta\hat{k}) = 1 \cdot 1 + 5 \cdot 3 + \alpha \cdot \beta = 0 \]
This simplifies to:
\[ 1 + 15 + \alpha \beta = 0 \implies \alpha \beta = -16 \tag{1} \]

### Step 3: Calculate \(\vec{B} \times \vec{C}\)
Now we need to calculate the cross product \(\vec{B} \times \vec{C}\):
\[ \vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & \beta \\ 1 & -2 & 3 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{B} \times \vec{C} = \hat{i}(3 \cdot 3 - \beta \cdot (-2)) - \hat{j}(1 \cdot 3 - \beta \cdot 1) + \hat{k}(1 \cdot (-2) - 3 \cdot 1) \]
This simplifies to:
\[ \vec{B} \times \vec{C} = \hat{i}(9 + 2\beta) - \hat{j}(3 - \beta) + \hat{k}(-2 - 3) \]
Thus:
\[ \vec{B} \times \vec{C} = (9 + 2\beta)\hat{i} + (-(3 - \beta))\hat{j} - 5\hat{k} \]

### Step 4: Find the magnitude of \(\vec{B} \times \vec{C}\)
We are given that the magnitude of \(\vec{B} \times \vec{C}\) is \(5\sqrt{3}\):
\[ |\vec{B} \times \vec{C}| = \sqrt{(9 + 2\beta)^2 + (3 - \beta)^2 + 25} = 5\sqrt{3} \]
Squaring both sides:
\[ (9 + 2\beta)^2 + (3 - \beta)^2 + 25 = 75 \]
This simplifies to:
\[ (9 + 2\beta)^2 + (3 - \beta)^2 = 50 \]

### Step 5: Expand and simplify
Expanding:
\[ (81 + 36\beta + 4\beta^2) + (9 - 6\beta + \beta^2) = 50 \]
Combining like terms:
\[ 5\beta^2 + 30\beta + 40 = 0 \]
Dividing by 5:
\[ \beta^2 + 6\beta + 8 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:
\[ \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm 2}{2} \]
This gives:
\[ \beta = -2 \quad \text{or} \quad \beta = -4 \]

### Step 7: Find corresponding \(\alpha\) values
Using equation (1):
1. If \(\beta = -2\):
\[ \alpha(-2) = -16 \implies \alpha = 8 \]
2. If \(\beta = -4\):
\[ \alpha(-4) = -16 \implies \alpha = 4 \]

### Step 8: Find the maximum value of \(|\vec{A}|^2\)
The magnitude squared of \(\vec{A}\) is:
\[ |\vec{A}|^2 = 1^2 + 5^2 + \alpha^2 = 1 + 25 + \alpha^2 = 26 + \alpha^2 \]
Calculating for both \(\alpha\) values:
1. For \(\alpha = 8\):
\[ |\vec{A}|^2 = 26 + 64 = 90 \]
2. For \(\alpha = 4\):
\[ |\vec{A}|^2 = 26 + 16 = 42 \]

### Conclusion
The maximum value of \(|\vec{A}|^2\) is:
\[ \boxed{90} \]