Find the shortest distance between the z-axis and the line, `x+y+2z-3=0,2x+3y+4z-4=0.`

Explanation

To find the shortest distance between the z-axis and the line defined by the intersection of the two planes given by the equations \(x + y + 2z - 3 = 0\) and \(2x + 3y + 4z - 4 = 0\), we can follow these steps:

### Step 1: Find the direction vector of the line of intersection of the two planes.
To find the direction vector of the line of intersection, we take the cross product of the normal vectors of the two planes.

The normal vector of the first plane \(x + y + 2z - 3 = 0\) is \(\mathbf{n_1} = (1, 1, 2)\).

The normal vector of the second plane \(2x + 3y + 4z - 4 = 0\) is \(\mathbf{n_2} = (2, 3, 4)\).
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