Find the equation of the line which passes through the point (-1,3,-2) and is perpendicular to the lines `(x)/(1)=(y)/(2)=(z)/(3)` and `vecr=(-2hati+hatj-hatk)+lamda(-3hati+2hatj+5hatk)`.

Answer

To find the equation of the line that passes through the point (-1, 3, -2) and is perpendicular to the given lines, we will follow these steps:

### Step 1: Identify the Direction Ratios of the Given Lines

The first line is given in symmetric form:
\[ \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \]
From this, we can extract the direction ratios (a1, b1, c1):
- \(a_1 = 1\)
- \(b_1 = 2\)
- \(c_1 = 3\)

The second line is given in vector form:
\[ \vec{r} = -2\hat{i} + \hat{j} - \hat{k} + \lambda(-3\hat{i} + 2\hat{j} + 5\hat{k}) \]
From this, the direction ratios (a2, b2, c2) are:
- \(a_2 = -3\)
- \(b_2 = 2\)
- \(c_2 = 5\)

### Step 2: Find the Cross Product of the Direction Ratios

To find a direction vector for the line we want, we need to take the cross product of the direction ratios of the two lines:
\[ \vec{b_1} = \langle 1, 2, 3 \rangle \quad \text{and} \quad \vec{b_2} = \langle -3, 2, 5 \rangle \]
The cross product \(\vec{b_1} \times \vec{b_2}\) can be calculated using the determinant:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix} \]

### Step 3: Calculate the Determinant

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & 3 \\ 2 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ -3 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 2 & 3 \\ 2 & 5 \end{vmatrix} = (2)(5) - (3)(2) = 10 - 6 = 4\)
2. \(\begin{vmatrix} 1 & 3 \\ -3 & 5 \end{vmatrix} = (1)(5) - (3)(-3) = 5 + 9 = 14\)
3. \(\begin{vmatrix} 1 & 2 \\ -3 & 2 \end{vmatrix} = (1)(2) - (2)(-3) = 2 + 6 = 8\)

Putting it all together:
\[ \vec{b_1} \times \vec{b_2} = 4\hat{i} - 14\hat{j} + 8\hat{k} = \langle 4, -14, 8 \rangle \]

### Step 4: Write the Equation of the Line

The line we want passes through the point (-1, 3, -2) and has direction ratios (4, -14, 8). The equation of the line in symmetric form is:
\[ \frac{x + 1}{4} = \frac{y - 3}{-14} = \frac{z + 2}{8} \]

### Final Answer

The equation of the line is:
\[ \frac{x + 1}{4} = \frac{y - 3}{-14} = \frac{z + 2}{8} \]