Let `veca = 2hati+lambda_1 hatj+3hatk`, `vec(b)=4hati+(3-lambda_2) hatj+6hatk` and `vec(c)=3hati+6hatj+(lambda_3-1)hatk` be three vectors such that `vec(b)=2 vec(a)` and `vec(a)` is perpendicular to `vec(c)` then a possible value of `((lambda)_1,(lambda)_2,(lambda)_3)` is:

(a) `(1,3,1)`
(b) `((-1/2),4,0)`
(c) `(1,5,1)`
(d) `((1/2), 4, -2)`

Options

A. (1, 5, 1)
B. (1, 3, 1)
C. `(1/2,4,-2)`
D. `(-1/2,4,0)`

Explanation

To solve the given problem, we have three vectors:

\[ \vec{a} = 2\hat{i} + \lambda_1 \hat{j} + 3\hat{k} \]
\[ \vec{b} = 4\hat{i} + (3 - \lambda_2) \hat{j} + 6\hat{k} \]
\[ \vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_3 - 1) \hat{k} \]

We know that \(\vec{b} = 2\vec{a}\) and \(\vec{a}\) is perpendicular to \(\vec{c}\).

### Step 1: Use the condition \(\vec{b} = 2\vec{a}\)

From the equation \(\vec{b} = 2\vec{a}\), we can substitute the expressions for \(\vec{a}\) and \(\vec{b}\):

\[ 4\hat{i} + (3 - \lambda_2) \hat{j} + 6\hat{k} = 2(2\hat{i} + \lambda_1 \hat{j} + 3\hat{k}) \]

Expanding the right side gives:

\[ 4\hat{i} + (2\lambda_1) \hat{j} + 6\hat{k} \]

Now, equate the components:

1. For the \(\hat{i}\) component:
\[ 4 = 4 \quad \text{(This is always true)} \]

2. For the \(\hat{j}\) component:
\[ 3 - \lambda_2 = 2\lambda_1 \quad \text{(Equation 1)} \]

3. For the \(\hat{k}\) component:
\[ 6 = 6 \quad \text{(This is also always true)} \]

### Step 2: Use the condition that \(\vec{a}\) is perpendicular to \(\vec{c}\)

The dot product of \(\vec{a}\) and \(\vec{c}\) must equal zero:

\[ \vec{a} \cdot \vec{c} = 0 \]

Calculating the dot product:

\[ (2\hat{i} + \lambda_1 \hat{j} + 3\hat{k}) \cdot (3\hat{i} + 6\hat{j} + (\lambda_3 - 1)\hat{k}) = 0 \]

Expanding this gives:

\[ 2 \cdot 3 + \lambda_1 \cdot 6 + 3(\lambda_3 - 1) = 0 \]

This simplifies to:

\[ 6 + 6\lambda_1 + 3\lambda_3 - 3 = 0 \]

Rearranging gives:

\[ 6\lambda_1 + 3\lambda_3 + 3 = 0 \]

Dividing through by 3:

\[ 2\lambda_1 + \lambda_3 + 1 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

Now we have two equations:

1. \(3 - \lambda_2 = 2\lambda_1\) (Equation 1)
2. \(2\lambda_1 + \lambda_3 + 1 = 0\) (Equation 2)

From Equation 1, we can express \(\lambda_2\):

\[ \lambda_2 = 3 - 2\lambda_1 \]

From Equation 2, we can express \(\lambda_3\):

\[ \lambda_3 = -2\lambda_1 - 1 \]

### Step 4: Check the options

Now we will check the provided options to find a consistent set of values for \((\lambda_1, \lambda_2, \lambda_3)\).

1. **Option (a)**: \((1, 3, 1)\)
- \(\lambda_1 = 1\)
- \(\lambda_2 = 3 - 2(1) = 1\) (not equal to 3)
- **Not valid.**

2. **Option (b)**: \((-1/2, 4, 0)\)
- \(\lambda_1 = -1/2\)
- \(\lambda_2 = 3 - 2(-1/2) = 3 + 1 = 4\) (valid)
- \(\lambda_3 = -2(-1/2) - 1 = 1 - 1 = 0\) (valid)
- **Valid option.**

3. **Option (c)**: \((1, 5, 1)\)
- \(\lambda_1 = 1\)
- \(\lambda_2 = 3 - 2(1) = 1\) (not equal to 5)
- **Not valid.**

4. **Option (d)**: \((1/2, 4, -2)\)
- \(\lambda_1 = 1/2\)
- \(\lambda_2 = 3 - 2(1/2) = 3 - 1 = 2\) (not equal to 4)
- **Not valid.**

### Conclusion

The only valid option is:

\[ \boxed{(-\frac{1}{2}, 4, 0)} \]