Question

I challenge whole brainly community. You can't solve it​

Answer 1

I can solve it

wait

your answer in 937

Let

p

=6

a

−4

b

+10

c

(P)

q

=−5

a

+3

b

+10

c

(Q)

r

=4

a

−6

b

−10

c

(R)

s

=2

b

+10

c

PQ=−11

a

+r

b

PR=−2

a

−2

b

−20

c

PS=−6

a

+6

b

[PQ PR PS]=0 for coplarality of points

[PQ PR PS=

∣

∣

∣

∣

∣

∣

∣

∣

−11

−2

−6

7

−2

6

0

−20

0

∣

∣

∣

∣

∣

∣

∣

∣

[

a

b

c

]

=20(−66+42)[

a

b

c

]

=−240[

a

b

c

]

[

a

b

c

]



=0→ not coplanar

So there points are not coplanar