TG EAMCET 2 May 2025 Shift 1 Paper

Let P1=(2,1,2) and P2=(1,2,1)
So, P1P2=P2−P1
‌⇒‌‌(1,2,1)−(2,1,2)
‌⇒‌‌(1−2,2−1,1−2
‌⇒‌‌(−1,1,−1)
The plane 2x−y+2z=1 has a normal vector n1=(2,−1,2).
Since, the required plane is perpendicular to 2x−y+2z=1, its normal vector n=(a,b,c) must be perpendicular to n1.
So, n must be perpendicular to P1P2.
∴n is parallel to the cross product of P1P2 and n1.
n‌=P1P2×n1=|

∧ i	∧ j	∧ k
−1	1	−1
2	−1	2

|
‌=(2−1)

∧

i

−(−2+2

∧

j

+(1−2

∧

k

‌=

∧

i

−

∧

k

=⟨1,0,−1⟩
Now, equation of the point (2,1,2) and the normal vector (1,0,−1) is
‌1(x−2+0(y−1)−1(z−2)=0
‌⇒‌‌x−2−z+2=0
‌⇒‌‌x−z=0
Comparing this with ax+by+cz+d=0, we have
‌a=1,b=0,c=−1,d=0
‌∴‌

a+b

c+d

=‌

1+0

−1+0

=−1