The equation of the given line is

x−2

2

=

y+1

4

=

z−2

2

... (1)
Any point on the given line is (3λ + 2, 4λ - 1 , 2λ + 2)
If this point lies on the given plane x – y + z – 5 = 0, then
3λ + 2 -(4λ - 1) + 2λ + 2 - 5 = 0
⇒ λ = 0
Putting λ = 0 in (3λ + 2, 4λ - 1 , 2λ + 2) , we get the point of intersection of the given line and the plane is (2, -1, 2).
Let θ be the angle between the given line and the plane
∴ sin θ =

→ a . → b

| → a || → b |

=

(3 ^ i +4 ^ j +2 ^ k ).( ^ i − ^ j + ^ k )

√32+42+22√12+12+12

=

3−4+2

√29√3

=

1

√87

⇒ θ = sin−1(

1

√87

)
Thus, the angle between the given line and the given plane is sin−1(

1

√87

)
OR
The equation of the given planes are

→

r

.

^

i

+2

^

j

+3

^

k

−4 = 0 ... (1)

→

r

.2

^

i

+

^

j

−

^

k

+5 = 0 ... (2)
The equation of the plane passing through the intersection of the planes (1) and (2) is
|

→

r

.

^

i

+2

^

j

+3

^

k

−4| + λ |

→

r

.2

^

i

+

^

j

−

^

k

+5| = 4 - 5λ ... (3)
Given that plane (3) is perpendicular to the plane

→

r

.5

^

i

+3

^

j

−6

^

k

+8 = 0
1 + 2λ × 5 + 2 + λ × 3 + 3 - λ × - 6 = 0
⇒ 19λ - 7 = 0
⇒ λ =

7

19

Putting λ =

7

19

in (3), we get
= 4 -

35

19

⇒

→

r

.(

33

19

^

i

+

45

19

^

j

+

50

19

^

k

) =

41

19

⇒

→

r

.33

^

i

+45

^

j

+50

^

k

= 41. This is the equation of the required plane.