Find the value of `k` if the straight line `2x+3y+4+k(6x-y+12)=0` is perpendicular to the line `7x+5y-4=0.`

Explanation

To find the value of \( k \) such that the line \( 2x + 3y + 4 + k(6x - y + 12) = 0 \) is perpendicular to the line \( 7x + 5y - 4 = 0 \), we can follow these steps:

### Step 1: Rewrite the first line
We start with the equation of the first line:
\[ 2x + 3y + 4 + k(6x - y + 12) = 0 \]
Expanding this, we get:
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