Distance between the two planes: `2x + 3y + 4z = 4`and `4x + 6y + 8z = 12`is(A) 2 units (B) 4 units (C) 8 units (D) `2/(sqrt(29))`units

Explanation

To find the distance between the two planes given by the equations \(2x + 3y + 4z = 4\) and \(4x + 6y + 8z = 12\), we can follow these steps:

### Step 1: Rewrite the equations of the planes in the standard form
The standard form of a plane equation is given by \(ax + by + cz + d = 0\).

For the first plane:
\[
2x + 3y + 4z - 4 = 0 \quad \text{(let's call this Plane 1)}
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