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. `vecx.vecd=-1`
B. `vecy.vecd= 1`
C. `vecz.vecd=0`
D. `vecr.vecd=0`, where `vecr=lambda vecx + mu vecy +deltavecz`

Explanation

To solve the problem step by step, we will analyze the given vectors and their relationships.

### Step 1: Identify the vectors
We have three vectors:
- \(\vec{a} = \hat{i} - \hat{j} + \hat{k}\)
- \(\vec{b} = 2\hat{i} + \hat{j}\)
- \(\vec{c} = 3\hat{j} - 2\hat{k}\)

### Step 2: Determine the vector \(\vec{d}\)
The vector \(\vec{d}\) is equally inclined to the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). This means that the angle between \(\vec{d}\) and each of these vectors is the same.

### Step 3: Use the property of coplanarity
Since \(\vec{d}\) is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) must be coplanar. For three vectors to be coplanar, the scalar triple product must be zero:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \]

### Step 4: Calculate the cross product \(\vec{b} \times \vec{c}\)
To find \(\vec{b} \times \vec{c}\):
\[ \vec{b} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, \quad \vec{c} = \begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix} \]
Using the determinant method:
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 0 & 3 & -2 \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i}(1 \cdot (-2) - 0 \cdot 3) - \hat{j}(2 \cdot (-2) - 0 \cdot 0) + \hat{k}(2 \cdot 3 - 1 \cdot 0) \]
\[ = -2\hat{i} + 4\hat{j} + 6\hat{k} \]

### Step 5: Calculate the dot product \(\vec{a} \cdot (\vec{b} \times \vec{c})\)
Now, calculate \(\vec{a} \cdot (\vec{b} \times \vec{c})\):
\[ \vec{a} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \quad \vec{b} \times \vec{c} = \begin{pmatrix} -2 \\ 4 \\ 6 \end{pmatrix} \]
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = 1 \cdot (-2) + (-1) \cdot 4 + 1 \cdot 6 = -2 - 4 + 6 = 0 \]
This confirms that \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar.

### Step 6: Analyze the vectors \(\vec{x}\), \(\vec{y}\), and \(\vec{z}\)
Let \(\vec{x}\), \(\vec{y}\), and \(\vec{z}\) be the vectors in the planes of \(\vec{a}, \vec{b}\); \(\vec{b}, \vec{c}\); and \(\vec{c}, \vec{a}\) respectively. Since \(\vec{d}\) is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we can conclude:
\[ \vec{x} \cdot \vec{d} = 0, \quad \vec{y} \cdot \vec{d} = 0, \quad \vec{z} \cdot \vec{d} = 0 \]

### Step 7: Formulate the vector \(\vec{r}\)
Let \(\vec{r} = \lambda \vec{x} + \mu \vec{y} + \delta \vec{z}\). Then:
\[ \vec{r} \cdot \vec{d} = \lambda (\vec{x} \cdot \vec{d}) + \mu (\vec{y} \cdot \vec{d}) + \delta (\vec{z} \cdot \vec{d}) = 0 \]

### Conclusion
From the above analysis, we can conclude:
- \( \vec{x} \cdot \vec{d} = 0 \)
- \( \vec{y} \cdot \vec{d} = 0 \)
- \( \vec{z} \cdot \vec{d} = 0 \)
- \( \vec{r} \cdot \vec{d} = 0 \)

Thus, the correct options are:
- \( z \cdot d = 0 \) (Option C)
- \( r \cdot d = 0 \) (Option D)