To find the equation of the lines passing through the point (2, 1, 3) 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 given in symmetric form:
\[
\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}
\]
From this, we can extract the direction ratios as \( (1, 2, 3) \).
The second line is given as:
\[
\frac{x}{-3} = \frac{y}{2} = \frac{z}{5}
\]
From this, the direction ratios are \( (-3, 2, 5) \).
### Step 2: Set Up the Equation of the Required Line
Let the direction ratios of the required line be \( (A, B, C) \). The line passes through the point \( (2, 1, 3) \), so we can write the equation of the line in symmetric form as:
\[
\frac{x - 2}{A} = \frac{y - 1}{B} = \frac{z - 3}{C}
\]
### Step 3: Use the Perpendicularity Condition
Since the required line is perpendicular to both given lines, we can use the condition for perpendicularity. For two lines to be perpendicular, the dot product of their direction ratios must equal zero.
1. For the first line:
\[
A \cdot 1 + B \cdot 2 + C \cdot 3 = 0 \quad \text{(Equation 1)}
\]
This simplifies to:
\[
A + 2B + 3C = 0
\]
2. For the second line:
\[
A \cdot (-3) + B \cdot 2 + C \cdot 5 = 0 \quad \text{(Equation 2)}
\]
This simplifies to:
\[
-3A + 2B + 5C = 0
\]
### Step 4: Solve the System of Equations
Now we have a system of two equations:
1. \( A + 2B + 3C = 0 \)
2. \( -3A + 2B + 5C = 0 \)
We can solve these equations simultaneously.
From Equation 1, we can express \( A \) in terms of \( B \) and \( C \):
\[
A = -2B - 3C
\]
Substituting this expression for \( A \) into Equation 2:
\[
-3(-2B - 3C) + 2B + 5C = 0
\]
This simplifies to:
\[
6B + 9C + 2B + 5C = 0
\]
Combining like terms gives:
\[
8B + 14C = 0
\]
From this, we can express \( B \) in terms of \( C \):
\[
B = -\frac{14}{8}C = -\frac{7}{4}C
\]
Now substituting \( B \) back into the expression for \( A \):
\[
A = -2\left(-\frac{7}{4}C\right) - 3C = \frac{14}{4}C - 3C = \frac{14}{4}C - \frac{12}{4}C = \frac{2}{4}C = \frac{1}{2}C
\]
### Step 5: Express Direction Ratios in Terms of a Parameter
Let \( C = k \). Then:
\[
A = \frac{1}{2}k, \quad B = -\frac{7}{4}k, \quad C = k
\]
### Step 6: Write the Final Equation
Substituting these values back into the symmetric form of the line:
\[
\frac{x - 2}{\frac{1}{2}k} = \frac{y - 1}{-\frac{7}{4}k} = \frac{z - 3}{k}
\]
To eliminate \( k \), we can multiply through by \( 4k \):
\[
4(x - 2) = 2(y - 1)(-7) = 4(z - 3)
\]
This gives us the final equation of the line:
\[
\frac{x - 2}{2} = \frac{y - 1}{-7} = \frac{z - 3}{4}
\]
