Three Dimensional Geometry Part 2

Let ⟨a,b,c⟩ be direction ratios of plane containing
lines‌

x

2

=‌

y

3

=‌

z

5

and ‌

x

3

=‌

y

7

=‌

z

8

.
∴‌‌2a+3b+5c=0‌‌...‌ (i) ‌
and 3a+7b+8c=0‌‌... (ii)
from eq. (i) and (ii) : ‌

a

24−35

=‌

b

15−16

=‌

c

14−9

∴‌‌ D. RS. of plane are <11,1,−5>
Let D⋅RS of plane P be <a1,b1,c1> then.
11a1+b1−5c1=0
and 9a1−b1−5c1=0
From eq. (iii) and (iv) :
‌

a1

−5−5

=‌

b1

−45+55

=‌

c1

−11−9

∴‌‌ D. A ‌5. of plane P are <1,−1,2>

Equation plane P is : 1(x−3)−1(y+4)+2(z−7)=0
⇒x−y+2z−21=0
Distance from point (2,−5,11) is d=‌

|2+5+22−2|

√6

∴d2=‌

32

3