A line passes through `(1, -1, 3)` and is perpendicular to the lines `vecr=(hat(i)+hat(j)-hat(k))+lambda(2hat(i)-2hat(j)+hat(k))` and `vecr=(2hat(i)-hat(j)-3hat(k))+mu(hat(i)+2hat(j)+2hat(k)).` Obtain its equation.

Answer

To find the equation of the line that passes through the point \( (1, -1, 3) \) and is perpendicular to the given lines, we will follow these steps:

### Step 1: Identify Direction Vectors of the Given Lines
The first line \( L_1 \) is given by:
\[ \vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \]
From this, we can identify the direction vector \( \vec{b} \) of line \( L_1 \):
\[ \vec{b} = 2\hat{i} - 2\hat{j} + \hat{k} \]

The second line \( L_2 \) is given by:
\[ \vec{r} = 2\hat{i} - \hat{j} - 3\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \]
From this, we can identify the direction vector \( \vec{d} \) of line \( L_2 \):
\[ \vec{d} = \hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 2: Find the Cross Product of the Direction Vectors
To find a direction vector for the line we want, we need to calculate the cross product \( \vec{b} \times \vec{d} \):
\[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{vmatrix} \]

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

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} -2 & 1 \\ 2 & 2 \end{vmatrix} = (-2)(2) - (1)(2) = -4 - 2 = -6 \)
2. \( \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = (2)(2) - (1)(1) = 4 - 1 = 3 \)
3. \( \begin{vmatrix} 2 & -2 \\ 1 & 2 \end{vmatrix} = (2)(2) - (-2)(1) = 4 + 2 = 6 \)

Putting it all together:
\[ \vec{b} \times \vec{d} = -6\hat{i} - 3\hat{j} + 6\hat{k} \]

### Step 3: Write the Equation of the Line
The line we are looking for passes through the point \( (1, -1, 3) \) and has a direction vector \( \vec{n} = -6\hat{i} - 3\hat{j} + 6\hat{k} \). The equation of the line can be expressed in vector form as:
\[ \vec{r} = \vec{P} + \alpha \vec{n} \]
where \( \vec{P} = \hat{i} - \hat{j} + 3\hat{k} \) and \( \alpha \) is a scalar parameter.

### Final Equation
Thus, the equation of the line can be written as:
\[ \vec{r} = (1, -1, 3) + \alpha (-6, -3, 6) \]