Find the equations of the line passing through the point (-1,3,-2) and perpendicular to each of the lines `(x)/(1) =(y)/(2)=(z)/(3) " and " (x+2)/(-3) =(y-1)/(2)=(z+1)/(5)`

Explanation

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

### Step 1: Identify the direction ratios of the given lines.
The first line is represented by the equation:
\[ \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \]
From this, we can extract the direction ratios as \( \mathbf{A} = (1, 2, 3) \).

The second line is represented by the equation:
\[ \frac{x + 2}{-3} = \frac{y - 1}{2} = \frac{z + 1}{5} \]
From this, we can extract the direction ratios as \( \mathbf{B} = (-3, 2, 5) \).

### Step 2: Compute the cross product of the direction ratios.
To find a direction vector that is perpendicular to both lines, we compute the cross product \( \mathbf{A} \times \mathbf{B} \).

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix} \]

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

Calculating the minors:
\[ = \mathbf{i} (2 \cdot 5 - 3 \cdot 2) - \mathbf{j} (1 \cdot 5 - 3 \cdot (-3)) + \mathbf{k} (1 \cdot 2 - 2 \cdot (-3)) \]
\[ = \mathbf{i} (10 - 6) - \mathbf{j} (5 + 9) + \mathbf{k} (2 + 6) \]
\[ = 4\mathbf{i} - 14\mathbf{j} + 8\mathbf{k} \]

Thus, the direction vector of the line we are looking for is \( \mathbf{D} = (4, -14, 8) \).

### Step 3: Write the equation of the line in parametric form.
The parametric equations of a line can be expressed as:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \]
where \( (x_1, y_1, z_1) \) is a point on the line and \( (a, b, c) \) are the direction ratios.

Substituting \( (x_1, y_1, z_1) = (-1, 3, -2) \) and \( (a, b, c) = (4, -14, 8) \):
\[ \frac{x + 1}{4} = \frac{y - 3}{-14} = \frac{z + 2}{8} \]

### Final Result
The equation of the line passing through the point (-1, 3, -2) and perpendicular to the given lines is:
\[ \frac{x + 1}{4} = \frac{y - 3}{-14} = \frac{z + 2}{8} \]

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