If lines `2x-3y+6=0` and `kx+2y=12=0` cut the coordinate axes in concyclic points, then the value of `|k|` is

Explanation

To solve the problem, we need to find the value of |k| given that the lines \(2x - 3y + 6 = 0\) and \(kx + 2y + 12 = 0\) cut the coordinate axes at concyclic points.

### Step 1: Find the intercepts of the first line \(2x - 3y + 6 = 0\)

To find the x-intercept, set \(y = 0\):
\[ 2x + 6 = 0 \implies 2x = -6 \implies x = -3 \]
So, the x-intercept is \((-3, 0)\).

To find the y-intercept, set \(x = 0\):
\[ -3y + 6 = 0 \implies -3y = -6 \implies y = 2 \]
So, the y-intercept is \((0, 2)\).

### Step 2: Find the intercepts of the second line \(kx + 2y + 12 = 0\)

To find the x-intercept, set \(y = 0\):
\[ kx + 12 = 0 \implies kx = -12 \implies x = -\frac{12}{k} \]
So, the x-intercept is \(\left(-\frac{12}{k}, 0\right)\).

To find the y-intercept, set \(x = 0\):
\[ 2y + 12 = 0 \implies 2y = -12 \implies y = -6 \]
So, the y-intercept is \((0, -6)\).

### Step 3: Identify the points

We have the following points:
- A: \((-3, 0)\)
- B: \((0, 2)\)
- C: \(\left(-\frac{12}{k}, 0\right)\)
- D: \((0, -6)\)

### Step 4: Use the condition for concyclic points

The points A, B, C, and D are concyclic if:
\[ OA \cdot OC = OB \cdot OD \]
Where \(O\) is the origin \((0, 0)\).

### Step 5: Calculate the distances

1. \(OA = \sqrt{(-3 - 0)^2 + (0 - 0)^2} = 3\)
2. \(OB = \sqrt{(0 - 0)^2 + (2 - 0)^2} = 2\)
3. \(OC = \sqrt{\left(-\frac{12}{k} - 0\right)^2 + (0 - 0)^2} = \frac{12}{|k|}\)
4. \(OD = \sqrt{(0 - 0)^2 + (-6 - 0)^2} = 6\)

### Step 6: Set up the equation

Substituting the distances into the concyclic condition:
\[ 3 \cdot \frac{12}{|k|} = 2 \cdot 6 \]
This simplifies to:
\[ \frac{36}{|k|} = 12 \]

### Step 7: Solve for \(|k|\)

Cross-multiplying gives:
\[ 36 = 12 |k| \implies |k| = \frac{36}{12} = 3 \]

### Final Answer

Thus, the value of \(|k|\) is:
\[ \boxed{3} \]