If a2 + b2 + c2 - ab - bc - ca = 0, then a ∶ b ∶ c is :

This question was previously asked in
Haryana CET Previous Year Paper (Held On: 6 Nov 2022 Shift 2)
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  1. ∶ 2 ∶ 1
  2. ∶ 1 ∶ 1
  3. ∶ 1 ∶ 2
  4. ∶ 1 ∶ 1
  5. Not attempted 

Answer (Detailed Solution Below)

Option 4 : ∶ 1 ∶ 1
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Haryana CET Full Test 1
100 Qs. 100 Marks 105 Mins

Detailed Solution

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Given :

a2 + b2 + c2 - ab - bc - ca = 0

Solution :

Here we have 3 variables and only one equation .

Note:  To solve these type of question we assume the value of any 2 variables.

Let assume that b = 1 and c = 1

⇒ a+ 1 + 1 - a - 1 - a = 0

⇒ a2 + 1 - 2a = 0

⇒ ( a - 1 )2 = 0                    [ ∵  ( A + B )2 = A2 + B2 + 2AB ]

Now a = 1

Now we have a = 1, b = 1 and c = 1

So a : b : c = 1 : 1 : 1

Hence the correct answer is "1 : 1 : 1".

Latest Haryana CET Updates

Last updated on Jul 10, 2025

->HSSC CET Exam Date 2025 is 26th and 27th July 2025.

->The Haryana HSSC CET 2025 Exam will be held for two days in 4 shifts.

->Earlier, Haryana CET Group C Notice for EWS Certificate was out. A valid format of EWS Certificate has been given in the Notice.

-> Haryana CET Group C Notification 2025 was out on 26th May 2025.

-> The minimum educational qualification to apply for the Common Eligibility Test is 10+2/equivalent 

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-> Aspirants must go through the Haryana CET Previous Years’ Papers to understand the need for the exam and prepare for the exam in the right direction.

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