Find the equation in vector and cartesian form of the line passing through the point :
(2,-1, 3) and perpendicular to the lines :
`vec(r) = (hat(i) + hat(j) - hat(k)) + lambda (2 hat(i) - 2 hat(j) + hat(k)) ` and
`vec(r) = (2 hat(i) - hat(j) - 3 hat(k) ) + mu (hat(i) + 2 hat(j) + 2 hat(k))` .

Answer

To find the equation of the line passing through the point \( P(2, -1, 3) \) and perpendicular to the given lines, we can follow these steps:

### Step 1: Identify the direction vectors of the given lines
The first line is given by:
\[ \vec{r_1} = \hat{i} + \hat{j} - \hat{k} + \lambda (2\hat{i} - 2\hat{j} + \hat{k}) \]
The direction vector of this line can be extracted as:
\[ \vec{A} = 2\hat{i} - 2\hat{j} + \hat{k} \]

The second line is given by:
\[ \vec{r_2} = (2\hat{i} - \hat{j} - 3\hat{k}) + \mu (\hat{i} + 2\hat{j} + 2\hat{k}) \]
The direction vector of this line can be extracted as:
\[ \vec{B} = \hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 2: Find the cross product of the direction vectors
To find a direction vector of the line that is perpendicular to both lines, we compute the cross product \( \vec{C} = \vec{A} \times \vec{B} \).

\[ \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{C} = \hat{i} \begin{vmatrix} -2 & 1 \\ 2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( (-2)(2) - (1)(2) = -4 - 2 = -6 \)
2. \( (2)(2) - (1)(1) = 4 - 1 = 3 \)
3. \( (2)(2) - (-2)(1) = 4 + 2 = 6 \)

Thus, we have:
\[ \vec{C} = -6\hat{i} - 3\hat{j} + 6\hat{k} \]

### Step 3: Write the vector equation of the line
The line passing through point \( P(2, -1, 3) \) with direction vector \( \vec{C} \) can be expressed as:
\[ \vec{r} = \vec{P} + \lambda \vec{C} \]
Where \( \vec{P} = 2\hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{C} = -6\hat{i} - 3\hat{j} + 6\hat{k} \).

Thus, the vector equation becomes:
\[ \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda (-6\hat{i} - 3\hat{j} + 6\hat{k}) \]

### Step 4: Convert to Cartesian form
To convert to Cartesian form, we express the vector equation in terms of \( x, y, z \):
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).

From the vector equation:
\[ x = 2 - 6\lambda \]
\[ y = -1 - 3\lambda \]
\[ z = 3 + 6\lambda \]

We can eliminate \( \lambda \) from these equations:
1. From \( x = 2 - 6\lambda \), we get \( \lambda = \frac{2 - x}{6} \).
2. Substituting \( \lambda \) in the equation for \( y \):
\[ y = -1 - 3\left(\frac{2 - x}{6}\right) = -1 - \frac{3(2 - x)}{6} = -1 - \frac{6 - 3x}{6} = -1 - 1 + \frac{x}{2} = \frac{x}{2} - 2 \]
3. Similarly, for \( z \):
\[ z = 3 + 6\left(\frac{2 - x}{6}\right) = 3 + 2 - x = 5 - x \]

Thus, the Cartesian equations are:
\[ \frac{x - 2}{-6} = \frac{y + 1}{-3} = \frac{z - 3}{6} \]

### Final Answer
The equation of the line in vector form is:
\[ \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda (-6\hat{i} - 3\hat{j} + 6\hat{k}) \]

And in Cartesian form:
\[ \frac{x - 2}{-6} = \frac{y + 1}{-3} = \frac{z - 3}{6} \]