Three Dimensional Geometry

Given equation of planeπis
ax+by+11z+d=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Now, the normal vector of π is
n1=a

∧

i

+b

∧

j

+11

∧

k

=⟨a,b,11⟩
Normal of first plane:
n2=2

∧

i

−3

∧

j

+1⋅

∧

k

=⟨2,−3,1⟩
Normal of second plane:
n3=3

∧

i

+

∧

j

−

∧

k

=⟨3,1,−1⟩
According to question, if plane π 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
‌‌

|d|

√22+52+112

=√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 =‌

−d

2

>0⇒d<0y-intercept
‌=−‌

d

5

>0⇒d<0
‌z‌-intercept ‌=‌

−d

11

>0⇒d<0
Hence, d=−30=−3×2×5=−3ab