The cartesian equation of a line is `6x+1=3y-2 = 3-2x`. Find its direction ratios.

Answer

To find the direction ratios of the line given by the Cartesian equation \(6x + 1 = 3y - 2 = 3 - 2z\), we will follow these steps:

### Step 1: Write the given equation in a standard form
The given equation is:
\[ 6x + 1 = 3y - 2 = 3 - 2z \]
We can rewrite this in a more manageable form by expressing each part separately:
\[ 6x + 1 = k, \quad 3y - 2 = k, \quad 3 - 2z = k \]
where \(k\) is a common parameter.

### Step 2: Rearrange each equation
From the first equation:
\[ 6x + 1 = k \implies 6x = k - 1 \implies x = \frac{k - 1}{6} \]
From the second equation:
\[ 3y - 2 = k \implies 3y = k + 2 \implies y = \frac{k + 2}{3} \]
From the third equation:
\[ 3 - 2z = k \implies 2z = 3 - k \implies z = \frac{3 - k}{2} \]

### Step 3: Express \(x\), \(y\), and \(z\) in terms of \(k\)
Now we have:
\[ x = \frac{k - 1}{6}, \quad y = \frac{k + 2}{3}, \quad z = \frac{3 - k}{2} \]

### Step 4: Eliminate \(k\) to find the direction ratios
To express the equations in a standard form, we can eliminate \(k\). We can set:
\[ \frac{x - \frac{-1}{6}}{1} = \frac{y - \frac{2}{3}}{2} = \frac{z - \frac{3}{2}}{-3} \]
This gives us the standard form of the line:
\[ \frac{x + \frac{1}{6}}{1} = \frac{y - \frac{2}{3}}{2} = \frac{z - \frac{3}{2}}{-3} \]

### Step 5: Identify the direction ratios
From the standard form, we can identify the direction ratios \(a\), \(b\), and \(c\):
- \(a = 1\)
- \(b = 2\)
- \(c = -3\)

Thus, the direction ratios of the line are:
\[ \text{Direction Ratios} = (1, 2, -3) \]

### Final Answer
The direction ratios of the line are \(1, 2, -3\).

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