(i) Find the vector equation of the line passing through (1,2,3) and parallel to the planes :
`vec(r) . (hati - hatj + 2 hatk ) = 5 and vec(r) . (3 hati + hatj + hatk)` = 6 .
(ii) Find the vector equation of the straight line passing through (1,2,3) and perpendicular to the plane :
`vec(r) . (hati + 2 hatj - 5 hatk ) + 9 = 0`.

Answer

To solve the given problem, we will break it down into two parts as per the question.

### Part (i): Find the vector equation of the line passing through (1,2,3) and parallel to the planes.

1. **Identify the position vector of the point**: The point given is (1, 2, 3). The position vector \( \vec{a} \) can be written as:
\[ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \]

2. **Find the normal vectors of the planes**: The equations of the planes are given as:
- Plane 1: \( \vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 5 \)
- Plane 2: \( \vec{r} \cdot (3\hat{i} + \hat{j} + \hat{k}) = 6 \)

The normal vector for Plane 1, \( \vec{n_1} \), is:
\[ \vec{n_1} = \hat{i} - \hat{j} + 2\hat{k} \]

The normal vector for Plane 2, \( \vec{n_2} \), is:
\[ \vec{n_2} = 3\hat{i} + \hat{j} + \hat{k} \]

3. **Calculate the direction vector of the line**: The line we are looking for is parallel to both planes, which means it is perpendicular to both normal vectors. To find a direction vector \( \vec{b} \) for the line, we take the cross product of \( \vec{n_1} \) and \( \vec{n_2} \):
\[ \vec{b} = \vec{n_1} \times \vec{n_2} \]

Calculating the cross product:
\[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} \]

Expanding this determinant:
\[ \vec{b} = \hat{i}((-1)(1) - (2)(1)) - \hat{j}((1)(1) - (2)(3)) + \hat{k}((1)(1) - (-1)(3)) \]
\[ = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) \]
\[ = -3\hat{i} + 5\hat{j} + 4\hat{k} \]

4. **Write the vector equation of the line**: The vector equation of the line can be expressed as:
\[ \vec{r} = \vec{a} + \lambda \vec{b} \]
Substituting \( \vec{a} \) and \( \vec{b} \):
\[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-3\hat{i} + 5\hat{j} + 4\hat{k}) \]

### Final Equation for Part (i):
\[ \vec{r} = (1 - 3\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (3 + 4\lambda)\hat{k} \]

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### Part (ii): Find the vector equation of the straight line passing through (1,2,3) and perpendicular to the plane.

1. **Identify the position vector of the point**: The point is again (1, 2, 3), so:
\[ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \]

2. **Find the normal vector of the plane**: The equation of the plane is given as:
\[ \vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0 \]
The normal vector \( \vec{n} \) of this plane is:
\[ \vec{n} = \hat{i} + 2\hat{j} - 5\hat{k} \]

3. **Write the vector equation of the line**: The line is perpendicular to the plane, which means its direction vector is the same as the normal vector of the plane. Thus, the vector equation of the line is:
\[ \vec{r} = \vec{a} + \mu \vec{n} \]
Substituting \( \vec{a} \) and \( \vec{n} \):
\[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(\hat{i} + 2\hat{j} - 5\hat{k}) \]

### Final Equation for Part (ii):
\[ \vec{r} = (1 + \mu)\hat{i} + (2 + 2\mu)\hat{j} + (3 - 5\mu)\hat{k} \]

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