Vectors `vec(a)=hat(i)+hat(j)+hat(k), vec(b)=hat(j)+3hat(k)` and `vec(c )=hat(i)-2hat(j)+hat(k)` are given. Find vector `vec(d)` if `vec(d)` is perpendicular to `vec(c )` and `vec(d).vec(a)=6, vec(d).vec(b)=11`.

Answer

To find the vector \(\vec{d}\) given the conditions that it is perpendicular to \(\vec{c}\) and satisfies the dot products \(\vec{d} \cdot \vec{a} = 6\) and \(\vec{d} \cdot \vec{b} = 11\), we can follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{j} + 3\hat{k} \]
\[ \vec{c} = \hat{i} - 2\hat{j} + \hat{k} \]
Assume:
\[ \vec{d} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 2: Use the condition for perpendicularity
Since \(\vec{d}\) is perpendicular to \(\vec{c}\), we have:
\[ \vec{d} \cdot \vec{c} = 0 \]
Calculating the dot product:
\[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = x(1) + y(-2) + z(1) = 0 \]
This simplifies to:
\[ x - 2y + z = 0 \quad \text{(Equation 1)} \]

### Step 3: Use the dot product with \(\vec{a}\)
From the condition \(\vec{d} \cdot \vec{a} = 6\):
\[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 \]
Calculating the dot product:
\[ x(1) + y(1) + z(1) = 6 \]
This simplifies to:
\[ x + y + z = 6 \quad \text{(Equation 2)} \]

### Step 4: Use the dot product with \(\vec{b}\)
From the condition \(\vec{d} \cdot \vec{b} = 11\):
\[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{j} + 3\hat{k}) = 11 \]
Calculating the dot product:
\[ x(0) + y(1) + z(3) = 11 \]
This simplifies to:
\[ y + 3z = 11 \quad \text{(Equation 3)} \]

### Step 5: Solve the system of equations
Now we have a system of three equations:
1. \(x - 2y + z = 0\) (Equation 1)
2. \(x + y + z = 6\) (Equation 2)
3. \(y + 3z = 11\) (Equation 3)

#### Step 5.1: Solve for \(y\) and \(z\)
From Equation 3:
\[ y = 11 - 3z \]

Substituting \(y\) into Equation 2:
\[ x + (11 - 3z) + z = 6 \]
This simplifies to:
\[ x + 11 - 2z = 6 \implies x - 2z = -5 \quad \text{(Equation 4)} \]

#### Step 5.2: Substitute \(y\) into Equation 1
Substituting \(y\) into Equation 1:
\[ x - 2(11 - 3z) + z = 0 \]
This simplifies to:
\[ x - 22 + 6z + z = 0 \implies x + 7z = 22 \quad \text{(Equation 5)} \]

### Step 6: Solve Equations 4 and 5
Now we have:
1. \(x - 2z = -5\) (Equation 4)
2. \(x + 7z = 22\) (Equation 5)

Subtract Equation 4 from Equation 5:
\[ (x + 7z) - (x - 2z) = 22 + 5 \]
This simplifies to:
\[ 9z = 27 \implies z = 3 \]

### Step 7: Find \(y\) and \(x\)
Substituting \(z = 3\) back into Equation 3:
\[ y + 3(3) = 11 \implies y + 9 = 11 \implies y = 2 \]

Substituting \(y = 2\) and \(z = 3\) into Equation 2:
\[ x + 2 + 3 = 6 \implies x + 5 = 6 \implies x = 1 \]

### Step 8: Write the final vector \(\vec{d}\)
Thus, we have:
\[ x = 1, \quad y = 2, \quad z = 3 \]
So, the vector \(\vec{d}\) is:
\[ \vec{d} = 1\hat{i} + 2\hat{j} + 3\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k} \]

### Final Answer:
\[ \vec{d} = \hat{i} + 2\hat{j} + 3\hat{k} \]