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
x−0
1
=
y−4
−1
=
z−5
2
=
−(0−4+10)
[12+(−1)2+22]
If (x,y,z) be the foot of perpendicular drawn from (x1,y1,z1) to the plane ax+by+cz+d=0. Then,