.Let `vec a=2hat i+hat j+hat k` and `vec b` and `vec c` be two nonzero vectors such that `|vec a+vec b+vec c|=|vec a+vec b-vec c|` and `vec b*vec c=0.`
Consider the following two statements :
(A) `|vec a+vec lamda c| ge |vec a|` for `lambda in R`
(B) `vec a` and `vec c` are always parallel
Then

Options

A. Both (A) and (B) are correct
B. Only (A) is correct
C. Only (B) is correct
D. Neither (A) nor (B) is correct

Answer

To solve the problem, we need to analyze the given conditions and statements step by step.

### Given:
1. \(\vec{a} = 2\hat{i} + \hat{j} + \hat{k}\)
2. \(|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|\)
3. \(\vec{b} \cdot \vec{c} = 0\) (indicating that \(\vec{b}\) and \(\vec{c}\) are orthogonal)

### Step 1: Analyze the first condition
We start with the equation:
\[ |\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}| \]
Squaring both sides, we have:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a} + \vec{b} - \vec{c}|^2 \]
Expanding both sides:
\[ (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = (\vec{a} + \vec{b} - \vec{c}) \cdot (\vec{a} + \vec{b} - \vec{c}) \]
This gives:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{a} \cdot \vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{c}) \]
Cancelling out common terms, we have:
\[ 2(\vec{b} \cdot \vec{c}) + 2(\vec{a} \cdot \vec{c}) = -2(\vec{b} \cdot \vec{c}) - 2(\vec{a} \cdot \vec{c}) \]
Since \(\vec{b} \cdot \vec{c} = 0\), we simplify to:
\[ 2(\vec{a} \cdot \vec{c}) = 0 \implies \vec{a} \cdot \vec{c} = 0 \]
This indicates that \(\vec{a}\) and \(\vec{c}\) are orthogonal.

### Step 2: Analyze Statement A
We need to check if:
\[ |\vec{a} + \lambda \vec{c}| \geq |\vec{a}| \]
for all \(\lambda \in \mathbb{R}\).

Calculating the left-hand side:
\[ |\vec{a} + \lambda \vec{c}|^2 = |\vec{a}|^2 + |\lambda \vec{c}|^2 + 2(\vec{a} \cdot (\lambda \vec{c})) \]
Since \(\vec{a} \cdot \vec{c} = 0\), we have:
\[ |\vec{a} + \lambda \vec{c}|^2 = |\vec{a}|^2 + \lambda^2 |\vec{c}|^2 \]
We want:
\[ |\vec{a}|^2 + \lambda^2 |\vec{c}|^2 \geq |\vec{a}|^2 \]
This simplifies to:
\[ \lambda^2 |\vec{c}|^2 \geq 0 \]
This is true for all \(\lambda \in \mathbb{R}\) since the square of any real number is non-negative.

### Step 3: Analyze Statement B
We need to check if \(\vec{a}\) and \(\vec{c}\) are always parallel. From our earlier calculation, we found that:
\[ \vec{a} \cdot \vec{c} = 0 \]
This means that \(\vec{a}\) and \(\vec{c}\) are orthogonal, not parallel. Therefore, Statement B is false.

### Conclusion
- Statement A is true: \(|\vec{a} + \lambda \vec{c}| \geq |\vec{a}|\) for all \(\lambda \in \mathbb{R}\).
- Statement B is false: \(\vec{a}\) and \(\vec{c}\) are not always parallel.

### Final Answer
Only Statement A is correct.

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