TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

We have, Equation of plane
r‌‌=b+c+x(a−b)+y(c+a)
‌‌=(1−x)b+(l+y)c+(x+y)a . . . (i)
And equation of line
r=a+t(b−c) . . . (ii)
From Eqs. (i) and (ii), we have
a+t(b−c)=(x+y)a+(1−x)b+(1+y)c
⇒‌‌x+y=1 . . . (iii)
1−x=t . . . (iv)
1+y=−t . . . (v)
Solving Eqs. (iii), (iv) and (v), we get
t=‌

−1

2

,x=‌

3

2

‌ and ‌y=‌

−1

2

∴r=a−‌

1

2

(b−c)
⇒‌‌r=a‌

1

2

b+‌

1

2

c=la+mb+nc
∴l=1,m‌

−1

2

‌ and ‌n=‌

1

2

Now, 3l+4m+2n
‌‌=3(1)+4(−‌

1

2

)+2(‌

1

2

)=3−2+1=2