Three Dimensional Geometry Part 2

Given, A⇒(a,−2a,3)
B⇒(0,4,5)
C⇒(0,−a,−1)
Equation of plane P⇒Ix+my+nz=0
As, C is foot of perpendicular from A to plane P. So, CA |N, where N is the normal vector to the plane.
CA=(a−0)

∧

i

+(−2a+a)

∧

j

+(3+1)

∧

k

=a

∧

i

−a

∧

j

+4

∧

k

Now, CA |N
So, ‌

a

l

=‌

−a

m

=‌

4

n

=λ
where λ is any real number.
P⇒(

a

λ

)x−(

a

λ

)y+(

4

λ

)z=0
P⇒ax−ay+4z=0
C lies on plane.
So, a⋅0−a(−a)+4(−1)=0
a2−4=0⇒a=±2
As per the question, a>0, so a=2
So, equation of plane P⇒2x−2y+4z=0
P⇒x−y+2z=0
Coordinates of D
If (x,y,z) be the foot of perpendicular drawn from (x1,y1,z1) to the plane ax+by+cz+d=0.
Then,
Here, (x,y,z)=(0,4,5)
⇒‌‌x−0=−(y−4)=‌

z−5

2

=‌

−6

6

∴‌‌x=−1,y=5,z=3
C=(0,−2,−1)⇒D=(−1,5,3)
∴‌‌CD=√(0+1)2+(−2−5)2+(−1−3)2
=√1+49+16
CD=√66