The position vector of a point `P` is `vecr=xhat(i)+yhat(j)+zhat(k),` where `x, y, zinN and veca=hat(i)+2hat(j)+hat(k).` If `vecr*veca=20` and the number of possible of `P` is `9lambda`, then the value of `lambda` is:

Options

A. a) `81`
B. b) `9`
C. c) `100`
D. d) `36`

Answer

To solve the problem, we need to find the value of \( \lambda \) given the conditions of the position vector \( \vec{r} \) and the vector \( \vec{a} \). Let's break it down step by step.

### Step 1: Write the vectors
The position vector of point \( P \) is given as:
\[ \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \]
where \( x, y, z \in \mathbb{N} \) (natural numbers).

The vector \( \vec{a} \) is given as:
\[ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \]

### Step 2: Set up the dot product equation
We know from the problem statement that:
\[ \vec{r} \cdot \vec{a} = 20 \]
Calculating the dot product:
\[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = x \cdot 1 + y \cdot 2 + z \cdot 1 = x + 2y + z \]
Thus, we have:
\[ x + 2y + z = 20 \]

### Step 3: Rearranging the equation
We can rearrange the equation to express \( x + z \):
\[ x + z = 20 - 2y \]

### Step 4: Determine the values of \( y \)
Since \( y \) is a natural number, we can find the possible values for \( y \):
- The right-hand side \( 20 - 2y \) must be non-negative, so:
\[ 20 - 2y \geq 0 \implies y \leq 10 \]
Thus, \( y \) can take values from \( 1 \) to \( 10 \).

### Step 5: Calculate possibilities for each \( y \)
For each value of \( y \), we will determine the corresponding values of \( x \) and \( z \):
1. If \( y = 1 \):
\[ x + z = 20 - 2(1) = 18 \]
Possible pairs \( (x, z) \) are \( (1, 17), (2, 16), \ldots, (17, 1) \) → 17 possibilities.

2. If \( y = 2 \):
\[ x + z = 20 - 2(2) = 16 \]
Possible pairs → 15 possibilities.

3. If \( y = 3 \):
\[ x + z = 20 - 2(3) = 14 \]
Possible pairs → 13 possibilities.

4. If \( y = 4 \):
\[ x + z = 20 - 2(4) = 12 \]
Possible pairs → 11 possibilities.

5. If \( y = 5 \):
\[ x + z = 20 - 2(5) = 10 \]
Possible pairs → 9 possibilities.

6. If \( y = 6 \):
\[ x + z = 20 - 2(6) = 8 \]
Possible pairs → 7 possibilities.

7. If \( y = 7 \):
\[ x + z = 20 - 2(7) = 6 \]
Possible pairs → 5 possibilities.

8. If \( y = 8 \):
\[ x + z = 20 - 2(8) = 4 \]
Possible pairs → 3 possibilities.

9. If \( y = 9 \):
\[ x + z = 20 - 2(9) = 2 \]
Possible pairs → 1 possibility.

10. If \( y = 10 \):
\[ x + z = 20 - 2(10) = 0 \]
No possibilities since \( x, z \) must be natural numbers.

### Step 6: Sum the possibilities
Now we sum the number of possibilities for \( y = 1 \) to \( y = 9 \):
\[ 17 + 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1 = 81 \]

### Step 7: Relate to \( 9\lambda \)
We know that the total number of possibilities is given by:
\[ 9\lambda = 81 \]
Thus, solving for \( \lambda \):
\[ \lambda = \frac{81}{9} = 9 \]

### Final Answer
The value of \( \lambda \) is:
\[ \boxed{9} \]