Question

2[3 x -2]+3[2y 1 4]=[0 9 z]​

Answer 1

Answer:

Consider the equation,

3

x+2

=

4

2y+3

=

5

3z+4

=λ

Therefore, any point on this line is of the form,

[3λ−2,

2

4λ−3

,

3

5λ−4

]

Now, the line from the point (−2,3,−4) is 3λ,

2

(4λ−9)

,

3

5λ+8

The equation of a plane is 4x+12y−3z+1=0

Thus, Direction ratio of normal is 4,12,−3

Therefore,

4⋅3λ+12[

2

(4λ−9)

]−3[(

3

5λ+8

)]=0

12λ+24λ−54−5λ−8=0

31λ=62

λ=2

Hence, the required coordinates is (4,5/2,2)

Hence, the distance between coordinates (4,5/2,2) and (2,3,−4) is 17/2unit