The equation of plane passing through
(1,0,0) is
a(x−1)+b(y−0)+c(z−0)=0.....(i)
since, plane also passing through
(0,2,0). ⇒a(0−1)+b(2−0)+c(0−0)=0
⇒−a+2b=0
⇒a=2b....(ii)
Given, distance from origin to plane (i)
= ⇒||= ⇒= ⇒= [from Eq.(ii)]
⇒14b=6√5b2+c2 Squaring on both sides;
⇒196b2=36(5b2+c2) ⇒196b2=180b2+36c2 ⇒16b2=36c2 ⇒4b=6c....(iii)
From Eq. (ii) and Eq. (iii),
2a=4b=6c ⇒== So, required equation of plane is,
6(x−1)+3(y−0)+2(z−0)=0 or
6x+3y+2z−6=0