Find the equation of straight line passing through (1,2,3) and parallel to `(-x-2)/1=(y+3)/7=(2z-6)/3`

Answer

To find the equation of the straight line passing through the point (1, 2, 3) and parallel to the line given by the equation \((-x-2)/1 = (y+3)/7 = (2z-6)/3\), we can follow these steps:

### Step 1: Identify the direction ratios of the given line
The given line can be rewritten in a more standard form. The equation \((-x-2)/1 = (y+3)/7 = (2z-6)/3\) can be interpreted as follows:

- The direction ratios of the line can be extracted from the coefficients of the equations.
- From \((-x-2)/1\), we can see that the direction ratio corresponding to \(x\) is \(-1\).
- From \((y+3)/7\), the direction ratio corresponding to \(y\) is \(7\).
- From \((2z-6)/3\), we can simplify it to \((z-3)/\frac{3}{2}\), giving a direction ratio for \(z\) as \(\frac{3}{2}\).

Thus, the direction ratios \(L, M, N\) of the line are:
- \(L = -1\)
- \(M = 7\)
- \(N = \frac{3}{2}\)

### Step 2: Write the equation of the line
The general equation of a line in space that passes through a point \((x_1, y_1, z_1)\) and has direction ratios \(L, M, N\) is given by:

\[ \frac{x - x_1}{L} = \frac{y - y_1}{M} = \frac{z - z_1}{N} \]

Substituting the point \((1, 2, 3)\) and the direction ratios we found:

\[ \frac{x - 1}{-1} = \frac{y - 2}{7} = \frac{z - 3}{\frac{3}{2}} \]

### Step 3: Simplify the equation
Now we can rewrite the equation:

1. For \(x\):
\[ x - 1 = -1 \cdot t \implies x = 1 - t \]

2. For \(y\):
\[ y - 2 = 7t \implies y = 2 + 7t \]

3. For \(z\):
\[ z - 3 = \frac{3}{2}t \implies z = 3 + \frac{3}{2}t \]

Thus, the parametric equations of the line can be expressed as:
\[ x = 1 - t, \quad y = 2 + 7t, \quad z = 3 + \frac{3}{2}t \]

### Final Equation
The final symmetric form of the equation of the line is:
\[ \frac{x - 1}{-1} = \frac{y - 2}{7} = \frac{z - 3}{\frac{3}{2}} \]