© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
487
Taking
ˆ
ˆ
ˆ
r
xi
y j
r= + +z k
, we get
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
) [(1 2 )
(1 3 )
(1 4 ) ] 6 5
xi
y j
z k
i
j
k
+
+
⋅
+ λ
+
+ λ
+
+ λ
= − λ
or
(1 + 2λ ) x + (1 + 3λ) y + (1 + 4λ) z = 6 – 5λ
or
(x + y + z – 6 ) + λ  (2x + 3y + 4 z + 5) = 0 (2)
Given that the plane passes through the point (1,1,1), it must satisfy (2), i e (1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0
or
λ = 3
14
Putting the values of  λ in (1), we get
3
9
6 ˆ
ˆ
ˆ
1
1
1
7
14
7
r
i
j
k








+
+
+
+
+
















r
 = 
15
6 14
−
or
10
23
13 ˆ
ˆ
ˆ
7
14
7
r
i
j
k


+
+




r
 = 69
14
or
ˆ
ˆ
ˆ
(20
23
26 )
r
i
j
k
⋅
+
+
r
 = 69
which is the required vector equation of the plane