The line , that is coplanar to the line `(x+3)/(-3)=(y-1)/1 =(z-5)/5` ,is

Options

A. `(x+1)/1 =(y-2)/2 =(z-5)/5`
B. `(x+1)/(-1) =(y-2)/2 =(z-5)/4`
C. `(x+1)/(-1) =(y-2)/2 =(z-5)/5`
D. `(x-1)/(-1) =(y-2)/2 =(z-5)/5`

Answer

To determine a line that is coplanar with the given line \((x+3)/(-3) = (y-1)/1 = (z-5)/5\), we can follow these steps:

### Step 1: Identify the Direction Ratios and a Point on the Given Line

The given line can be expressed in parametric form. From the equation:

\[ \frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5} = t \]

We can derive the parametric equations:

\[ x = -3t - 3, \quad y = t + 1, \quad z = 5t + 5 \]

From this, we can identify the direction ratios of the line as \((-3, 1, 5)\) and a point on the line when \(t = 0\) is \((-3, 1, 5)\).

### Step 2: Choose a Point and Direction Ratios for the New Line

Let’s assume we have another line that we want to check for coplanarity. For example, we can take a point \(A(1, 2, 3)\) and direction ratios \(d_1(2, -1, 1)\).

### Step 3: Check for Coplanarity

To check if two lines are coplanar, we can use the scalar triple product. The lines will be coplanar if the scalar triple product of the direction ratios of the two lines and the vector connecting the two points is zero.

1. **Direction Ratios of the Given Line**: \(d_1 = (-3, 1, 5)\)
2. **Direction Ratios of the New Line**: \(d_2 = (2, -1, 1)\)
3. **Vector between Points**: \(A(-3, 1, 5)\) to \(B(1, 2, 3)\) gives us the vector \(AB = (1 - (-3), 2 - 1, 3 - 5) = (4, 1, -2)\).

### Step 4: Calculate the Scalar Triple Product

The scalar triple product is given by the determinant of the matrix formed by the three vectors:

\[ \begin{vmatrix} -3 & 1 & 5 \\ 2 & -1 & 1 \\ 4 & 1 & -2 \end{vmatrix} \]

Calculating this determinant:

\[ = -3 \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} - 1 \begin{vmatrix} 2 & 1 \\ 4 & -2 \end{vmatrix} + 5 \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (1)(1) = 2 - 1 = 1\)

2. \(\begin{vmatrix} 2 & 1 \\ 4 & -2 \end{vmatrix} = (2)(-2) - (1)(4) = -4 - 4 = -8\)

3. \(\begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} = (2)(1) - (-1)(4) = 2 + 4 = 6\)

Putting it all together:

\[ = -3(1) - 1(-8) + 5(6) = -3 + 8 + 30 = 35 \]

Since the scalar triple product is not zero, the lines are not coplanar.

### Conclusion

Thus, the line that is coplanar to the given line can be determined by ensuring that the scalar triple product of the direction ratios and the connecting vector is zero.