If the straight lines are coplanar find the real values of m. (x - 1)/1 = (y - 2)/2 = (z - 3)/(m ^ 2) (x - 3)/1 = (y - 2)/(m ^ 2) = (z - 1)/2

To find the real values of \( m \) such that the given straight lines are coplanar, we need to use the condition for coplanarity of two lines. The lines in symmetric form are given by:

1. \(\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{m^2}\)
2. \(\frac{x - 3}{1} = \frac{y - 2}{m^2} = \frac{z - 1}{2}\)

The condition for the coplanarity of two lines given in symmetric form:

\[
\text{Lines } \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \text{ and } \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2} \text{ are coplanar if the determinant:}
\]

\[
\begin{vmatrix}
x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix} = 0
\]

### Step 1: Identify the parameters
For the first line:
- Point \( P_1 \) is \((1, 2, 3)\)
- Direction ratios: \( a_1 = 1 \), \( b_1 = 2 \), \( c_1 = m^2 \)

For the second line:
- Point \( P_2 \) is \((3, 2, 1)\)
- Direction ratios: \( a_2 = 1 \), \( b_2 = m^2 \), \( c_2 = 2 \)

### Step 2: Apply the coplanarity condition

The vector joining the points \( P_1 \) and \( P_2 \) is:

\[
\overrightarrow{P_1P_2} = (3 - 1, 2 - 2, 1 - 3) = (2, 0, -2)
\]

Now, apply the coplanarity condition:

\[
\begin{vmatrix}
2 & 0 & -2 \\
1 & 2 & m^2 \\
1 & m^2 & 2
\end{vmatrix} = 0
\]

### Step 3: Calculate the determinant

Expand the determinant:

\[
= 2 \left(2 \cdot 2 - m^2 \cdot m^2\right) - 0 \cdot (1 \cdot 2 - m^2 \cdot 1) - 2 \left(1 \cdot m^2 - 2 \cdot 1\right)
\]

\[
= 2 \left(4 - m^4\right) - 2 \left(m^2 - 2\right)
\]

\[
= 8 - 2m^4 - 2m^2 + 4
\]

\[
= -2m^4 - 2m^2 + 12
\]

Set the determinant to zero:

\[
-2m^4 - 2m^2 + 12 = 0
\]

### Step 4: Solve the equation

Divide the entire equation by \(-2\):

\[
m^4 + m^2 - 6 = 0
\]

Let \( t = m^2 \). The equation becomes:

\[
t^2 + t - 6 = 0
\]

Factorizing the quadratic:

\[
(t - 2)(t + 3) = 0
\]

So, \( t = 2 \) or \( t = -3 \).

Since \( t = m^2 \), and \( m^2 \) cannot be negative:

\[
m^2 = 2 \Rightarrow m = \pm \sqrt{2}
\]

### Final Answer:
The real values of \( m \) for which the lines are coplanar are \( m = \sqrt{2} \) and \( m = -\sqrt{2} \).

Would you like a detailed explanation of any step? Here are 5 related questions:

1. How do you determine the direction ratios of a line given in symmetric form?
2. What is the general condition for two lines to be coplanar?
3. Can we extend this method to 3D planes intersecting lines?
4. How do we solve higher-degree polynomials like quartic equations?
5. What are some applications of coplanarity in geometry?

**Tip:** When working with determinants, always simplify the matrix first to avoid errors in expansion.

Discover how to find real values of m ensuring coplanarity of lines using vectors and determinants. Step-by-step explanation with formulas and solved example. Suitable for advanced high school students.

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