Let the position vectors of the three points be,

→

a

=

^

i

+

^

j

−2

^

k

,

→

b

= 2

^

i

−

^

j

+

^

k

and

→

c

=

^

i

+2

^

j

+

^

k

.
So, the equation of the plane passing through the points

→

a

,

→

b

and

→

c

is
(

→

r

−

→

a

).[(

→

b

−

→

c

)×(

→

c

−

→

a

)] = 0
⇒ [

→

r

−

^

i

+j^→+2

^

k

] . [

^

i

−3

^

j

×

^

j

+3

^

k

] = 0
⇒ [

→

r

−

^

i

+j^→+2

^

k

] .

^

k

−3

^

j

−9

^

i

= 0
⇒

→

r

.(9

^

i

+3

^

j

−

^

k

) = 14 ... (1)
So, the vector equation of the required plane is

→

r

.(9

^

i

+3

^

j

−

^

k

) = 14
The equation of the given line is

→

r

= 3

^

i

−

^

j

−

^

k

+λ(2

^

i

−2

^

j

+

^

k

)
Position vector of any point on the given line is

→

r

= 3 + 2λ

^

i

+ (- 1 - 2λ)

^

j

+ (- 1 + λ)

^

k

... (2)
The point (2) lies on plane (1) if,
= 14
⇒ 9 (3) + 2λ + 3 - 1 - 2λ - (-1) + λ = 14
⇒ 11λ + 25 = 14
⇒ λ = - 1
Putting λ = - 1 in (2), we have

→

r

= (3 + 2λ)

^

i

+ (- 1 - 2λ)

^

j

+ (- 1 + λ)

^

k

= (3 + 2 - 1)

^

i

+ (- 1 - 2 - 1)

^

j

+ (- 1 + (- 1))

^

k

=

^

i

+

^

j

−2

^

k

Thus, the position vector of the point of intersection of the given line and plane (1) is

^

i

+

^

j

−2

^

k

and its co-ordinates are 1, 1, - 2 .