JEE Main 24 Feb 2021 Shift 1 Solved Paper

Given, 3x−2y−kz=10
2x−4y−2z‌‌=6
x+2y−z‌‌=5m
Condition for inconsistency ⇒∆=0 and atleast one of the ∆x′∆y,∆z is non-zero.
‌ Now, ‌∆‌‌=|

3	−2	−k
2	−4	−2
1	2	−1

|
‌‌=3(4+4)+2(−2+2)−k(4+4)
‌‌=24−8k
Now, ∆=0
∴‌‌24−8k=0
⇒‌‌8k=24⇒k=‌

24

8

=3
∆x=|

10	−2	−k
6	−4	−2
5m	2	−1

|
=10(4+4)+2(−6+10m)−3(12+20m)
=80−12+20m−36−60m
=32−40m=8(4−5m)
∆y=|

3	10	−k
2	6	−2
1	5m	−1

|
=3(−6+10m)−10(−2+2)−3(10m−6)
=−18+30m−30m+18
=0 ∆z=|

3	−2	10
2	−4	6
1	2	5m

|
‌‌=3(−20m−12)+2(10m−6)+10(4+4)
‌‌=−60m−36+20m−12+80
‌‌=−40m+32=8(4−5m)
Here, either ∆x or ∆z≠0
‌⇒‌8(4−5m)‌≠0
⇒‌m≠4∕5
Hence, the required values are k=3;m≠‌

4

5

.