If $d_1, d_2, d_3$ denote the distances of the plane 2x - 3y + 4z = 0 from the planes 2x - 3y + 4z + 6 = 0, 4x - 6y + 8z + 3 = 0 and 2x - 3y + 4z - 6 = 0 respectively, then
Answer & explanation
Correct answer: option 3
We have,
$d_1 = \frac{|6-2|}{\sqrt{2^2+(-3)^2+4^2}}, d_2=\frac{|3/2-2|}{\sqrt{2^2+(-3)^2 +4^2}}$
and $d_3 = \frac{|-6-2|}{\sqrt{2^2+(-3)^2+4^2}}$
$⇒d_1 = \frac{4}{\sqrt{29}}, d_2= \frac{1}{2\sqrt{29}}$ and $d_3=\frac{8}{\sqrt{29}}$
Clearly, $8d_2 = d_1$