If the distance between planes 4x-2y-4z+1=0 and 4x-2y-4z+d=0 is 7, then d is:

Question

If the distance between planes $4x-2y-4z+1=0$ and $4x-2y-4z+d=0$ is 7, then $d$ is:

Accepted Answer

$$\frac{|d-1|}{\sqrt{16+4+16}}=7\implies|d-1|=42\implies d=43\text{ or }-41.$$

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