Let `vecp=hati+hatj+hatk` are `vecr` be a variable vector such that `vecr. hati`, `vecr. hatj` and `vecr. hatk` are even natural numbers. If `vecr.vecp le 20`, then the numbers. If `vecr.vecple20` then the number of values of `vecr` is

Options

A. 20
B. 60
C. 75
D. 120

Answer

To solve the problem, we need to analyze the given information step by step.

### Step 1: Define the vectors
Let \(\vec{p} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\), where \(x\), \(y\), and \(z\) are even natural numbers.

### Step 2: Set up the dot product
The dot product \(\vec{r} \cdot \vec{p}\) can be calculated as follows:
\[ \vec{r} \cdot \vec{p} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = x + y + z \]
We are given that:
\[ x + y + z \leq 20 \]

### Step 3: Express \(x\), \(y\), and \(z\) as even natural numbers
Since \(x\), \(y\), and \(z\) are even natural numbers, we can express them as:
\[ x = 2k_1, \quad y = 2k_2, \quad z = 2k_3 \]
where \(k_1\), \(k_2\), and \(k_3\) are natural numbers (i.e., \(k_1, k_2, k_3 \geq 1\)).

### Step 4: Substitute into the inequality
Substituting these into the inequality gives:
\[ 2k_1 + 2k_2 + 2k_3 \leq 20 \]
Dividing the entire inequality by 2:
\[ k_1 + k_2 + k_3 \leq 10 \]

### Step 5: Introduce a dummy variable
To convert the inequality into an equation, we introduce a dummy variable \(w\):
\[ k_1 + k_2 + k_3 + w = 10 \]
where \(w \geq 0\).

### Step 6: Change variables
Since \(k_1\), \(k_2\), and \(k_3\) must be at least 1, we can redefine them:
\[ k_1' = k_1 - 1, \quad k_2' = k_2 - 1, \quad k_3' = k_3 - 1 \]
where \(k_1', k_2', k_3' \geq 0\). Thus, we can rewrite the equation as:
\[ (k_1' + 1) + (k_2' + 1) + (k_3' + 1) + w = 10 \]
which simplifies to:
\[ k_1' + k_2' + k_3' + w = 7 \]

### Step 7: Count the non-negative integer solutions
We need to find the number of non-negative integer solutions to the equation:
\[ k_1' + k_2' + k_3' + w = 7 \]
This is a classic "stars and bars" problem, where the number of solutions is given by:
\[ \binom{n + r - 1}{r - 1} \]
where \(n\) is the total (7) and \(r\) is the number of variables (4).

### Step 8: Calculate the combinations
Thus, we have:
\[ \binom{7 + 4 - 1}{4 - 1} = \binom{10}{3} \]
Calculating \(\binom{10}{3}\):
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 \]

### Final Answer
The number of values of \(\vec{r}\) such that \(\vec{r} \cdot \vec{p} \leq 20\) is **120**.