Prove that the lines `(x+1)/3=(y+3)/5=(z+5)/7a n d(x-2)/1=(y-4)/4=(z-6)/7` are coplanar . Also, find the plane containing these two lines.

Explanation

To prove that the lines \(\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}\) and \(\frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}\) are coplanar, we can use the determinant method.

### Step 1: Identify the points and direction ratios of the lines
For the first line \(L_1: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}\):
- Point \(A(-1, -3, -5)\)
- Direction ratios: \(l_1 = 3\), \(m_1 = 5\), \(n_1 = 7\)

For the second line \(L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}\):
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