Let `vec(A)=4hat(i)+5hat(j)-hat(k), vec(b)=hat(i)-4hat(j)+5hat(k)` and `vec(c) =3hat(i)+hat(j)-hat(k)`. Find a vector `vec(d)` which is perpendicular to both `vec(a)` and `vec(b)`, and is such that `vec(d). Vec(c)=21`.

Explanation

To find a vector \(\vec{d}\) that is perpendicular to both \(\vec{a}\) and \(\vec{b}\), and satisfies the condition \(\vec{d} \cdot \vec{c} = 21\), we can follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = 4\hat{i} + 5\hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - 4\hat{j} + 5\hat{k} \]
\[ \vec{c} = 3\hat{i} + \hat{j} - \hat{k} \]
Let \(\vec{d} = x\hat{i} + y\hat{j} + z\hat{k}\).

### Step 2: Set up the equations for perpendicularity
Since \(\vec{d}\) is perpendicular to \(\vec{a}\):
\[ \vec{d} \cdot \vec{a} = 0 \]
This gives us:
\[ x(4) + y(5) + z(-1) = 0 \quad \text{(Equation 1)} \]
or
\[ 4x + 5y - z = 0 \]

Since \(\vec{d}\) is also perpendicular to \(\vec{b}\):
\[ \vec{d} \cdot \vec{b} = 0 \]
This gives us:
\[ x(1) + y(-4) + z(5) = 0 \quad \text{(Equation 2)} \]
or
\[ x - 4y + 5z = 0 \]

### Step 3: Solve the system of equations
We now have two equations:
1. \(4x + 5y - z = 0\)
2. \(x - 4y + 5z = 0\)

From Equation 1, we can express \(z\) in terms of \(x\) and \(y\):
\[ z = 4x + 5y \]

Substituting \(z\) into Equation 2:
\[ x - 4y + 5(4x + 5y) = 0 \]
Expanding this:
\[ x - 4y + 20x + 25y = 0 \]
Combining like terms:
\[ 21x + 21y = 0 \]
This simplifies to:
\[ x + y = 0 \quad \Rightarrow \quad y = -x \]

### Step 4: Substitute back to find \(z\)
Substituting \(y = -x\) into the expression for \(z\):
\[ z = 4x + 5(-x) = 4x - 5x = -x \]

### Step 5: Express \(\vec{d}\) in terms of \(x\)
Thus, we have:
\[ \vec{d} = x\hat{i} - x\hat{j} - x\hat{k} = x(\hat{i} - \hat{j} - \hat{k}) \]

### Step 6: Use the condition \(\vec{d} \cdot \vec{c} = 21\)
Now we need to satisfy:
\[ \vec{d} \cdot \vec{c} = 21 \]
Calculating \(\vec{d} \cdot \vec{c}\):
\[ \vec{d} \cdot \vec{c} = x(\hat{i} - \hat{j} - \hat{k}) \cdot (3\hat{i} + \hat{j} - \hat{k}) \]
Calculating the dot product:
\[ = x(3 - 1 + 1) = x(3) \]
Setting this equal to 21:
\[ 3x = 21 \quad \Rightarrow \quad x = 7 \]

### Step 7: Find the components of \(\vec{d}\)
Substituting \(x = 7\):
\[ y = -7, \quad z = -7 \]
Thus:
\[ \vec{d} = 7\hat{i} - 7\hat{j} - 7\hat{k} \]

### Final Result
The vector \(\vec{d}\) is:
\[ \vec{d} = 7\hat{i} - 7\hat{j} - 7\hat{k} \]