Three Dimensional Geometry Part 2

Let L1⇒‌

x−3

1

=‌

y−3

2

=‌

z−4

2

=u (say)
⇒ Direction ratios of ‌‌L1=1,2,2
L2⇒‌

x−3

2

=‌

y−3

2

=‌

z−2

1

=v‌ (say) ‌
Direction ratios of L2=2,2,1
Line L passing through origin is perpendicular to L1 and L2.
Hence, direction ratios of L is parallel to (L1×L2).
⇒‌‌(−2,3,−2)
Equation of L⇒‌

x

2

=‌

y

−3

=‌

z

2

=λ (say)
Solve L and L1, we get
(2λ,−3λ,2λ)=(µ+3,2µ−1,2µ+4)
Gives, λ=1,µ=−1
So, intersection point P(2,−3,2).
Let Q(2v+3,2v+3,v+2) be required point on L2.
Now, PQ=√17 (given)
‌ Now, ‌PQ=√17‌ (given) ‌
PQ‌‌=√(2v+1)2+(2v+6)2+(v)2
‌‌=√17
⇒(2v+1)2‌+(2v+6)2+v2=17‌‌‌ (squaring on both sides) ‌
⇒‌‌9v2+28v+20=0
On solving, we get v=−2 (rejected), ‌

−10

9

(accepted)
∴Q is (‌

7

9

,‌

7

9

,‌

8

9

)
∴18(a+b+c)=18(‌

7

9

+‌

7

9

+‌

8

9

)=44