A vector `vecd` is equally inclined to three vectors `veca=hati-hatj+hatk,vecb=2hati+hatj and vecc=3hatj-2hatk.` Let `vecx,vecy and vecz` be three vectors in the plane of `veca,vecb;vecb,vec;vecc,veca,` respectively. Then

Options

A. (a) `vecz.vecd =0`
B. (b) `vecx.vecd=1`
C. (c)`vecy.vecd= 32`
D. (d) `vecr.vecd= 0 , " where " vecr= lambdavecx+ muvecy+ gammavecz`

Explanation

To solve the problem step-by-step, we need to find the vector \(\vec{d}\) that is equally inclined to the three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

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

### Step 2: Set up the conditions for equal inclination
A vector \(\vec{d}\) is equally inclined to vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) if the angles between \(\vec{d}\) and each of these vectors are the same. This can be expressed using the dot product:
\[ \vec{d} \cdot \vec{a} = k_1, \quad \vec{d} \cdot \vec{b} = k_2, \quad \vec{d} \cdot \vec{c} = k_3 \]
where \(k_1 = k_2 = k_3\).

### Step 3: Express \(\vec{d}\) in terms of its components
Let:
\[ \vec{d} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 4: Write the dot products
Now, we can write the dot products:
1. \(\vec{d} \cdot \vec{a} = x(1) + y(-1) + z(1) = x - y + z\)
2. \(\vec{d} \cdot \vec{b} = x(2) + y(1) + z(0) = 2x + y\)
3. \(\vec{d} \cdot \vec{c} = x(0) + y(3) + z(-2) = 3y - 2z\)

### Step 5: Set the dot products equal to each other
From equal inclination, we have:
\[ x - y + z = 2x + y = 3y - 2z \]

### Step 6: Solve the equations
1. From \(x - y + z = 2x + y\):
\[ -y + z = x + y \implies z = 2y + x \]

2. From \(2x + y = 3y - 2z\):
Substitute \(z\) from the previous equation:
\[ 2x + y = 3y - 2(2y + x) \]
Simplifying:
\[ 2x + y = 3y - 4y - 2x \implies 4x + 2y = 0 \implies 2x + y = 0 \implies y = -2x \]

### Step 7: Substitute \(y\) back to find \(z\)
Substituting \(y = -2x\) into \(z = 2y + x\):
\[ z = 2(-2x) + x = -4x + x = -3x \]

### Step 8: Express \(\vec{d}\)
Thus, we can express \(\vec{d}\) in terms of \(x\):
\[ \vec{d} = x\hat{i} - 2x\hat{j} - 3x\hat{k} = x(\hat{i} - 2\hat{j} - 3\hat{k}) \]

### Step 9: Conclusion
The vector \(\vec{d}\) can be expressed as:
\[ \vec{d} = k(\hat{i} - 2\hat{j} - 3\hat{k}), \text{ where } k \text{ is a scalar.} \]