Question

3 x3 ܙܘܪ ; + 1 + X ܕ 6 (6) 3 (et) 2 (5), 1 () 0​

Answer 1

Answer:

Let points (−2,3,5),(1,2,3) and (7,0,−1) be denoted by P,Q and R respectively.

Points P,Q and R are collinear if they lie on a line.

PQ=  

(1+2)  

2

+(2−3)  

2

+(3−5)  

2

 

​

 

=  

(3)  

2

+(−1)  

2

+(−2)  

2

 

​

 

=  

9+1+4

​

 

=  

14

​

 

QR=  

(7−1)  

2

+(0−2)  

2

+(−1−3)  

2

 

​

 

=  

(6)  

2

+(−2)  

2

+(−4)  

2

 

​

 

=  

36+4+16

​

 

=  

56

​

 

=2  

14

​

 

PR=  

(7+2)  

2

+(0−3)  

2

+(−1−5)  

2

 

​

 

=  

(9)  

2

+(−3)  

2

+(−6)  

2

 

​

 

=  

81+9+36

​

 

=  

126

​

 

=3  

14

​

 

Here PQ+QR=  

14

​

+2  

14

​

 

=3  

14

​

 

=PR

⇒PQ+QR=PR

Step-by-step explanation: