Question

The distance between the parallel planes $4x–6y+2z+k=0$ and $2x–3y+z+1=0$ is $\sqrt{14}$, then $k=$

• A. $–13,15$
• B. $-15,12$
• C. $–26,30$
• D. $–30,26$

Answer 1

The correct option is C $–26,30$

$\frac{\mid 2-k\mid }{\sqrt{16+36+4}}=\sqrt{14}\Rightarrow \mid 2-k\mid =28\Rightarrow k=-26,30$