Given equation of plane
Ï€is
ax+by+11z+d=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)Now, the normal vector of
Ï€ is
n1=a+b+11=⟨a,b,11⟩Normal of first plane:
n2=2−3+1⋅=⟨2,−3,1⟩Normal of second plane:
n3=3+−=⟨3,1,−1⟩According to question, if plane
Ï€ is perpendicular to both then its normal vector is perpendicular to the normals of the two planes.
So,
n1⋅n2=0⇒2a−3b+11=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)n1⋅n3=0⇒3a+b−11=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)On solving Eqs. (ii) and (iii) we get,
‌‌‌a=2‌ and ‌b=11−3a=11−6=5‌∴<a,b,11>=<2,5,11> ∴ Equation of plane will become,
2x+5y+11z+d=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(iv) Now, distance from origin
(0,0,0) to the plane (iv) is
‌‌=√6⇒|d|=√6+√150=30⇒d=±30(given)
As question demand is all intercepts (i.e.
x,y×z intercepts) are positive.
So,
d must be negative because
x-intercept
=‌>0⇒d<0y-intercept
‌=−‌>0⇒d<0‌z‌-intercept ‌=‌>0⇒d<0Hence,
d=−30=−3×2×5=−3ab