AP EAPCET 05-Jul-2022 Shift 1 Paper

The equation of plane passes through the point (1,0,1) is
a(x−1)+b(y−0)+c(z−1)=0.......(i)
If this plane is perpendicular to the planes 2x+3y−z=2 and x−y+2z=1
Then, 2a+3b−c=2 and a−b+2c=1
‌⇒‌

a

6−1

=‌

b

−1−4

=‌

c

2(−1)−3

‌⇒‌

a

5

=‌

b

−3

=‌

c

−5

=k
Let a=5k,b=−3k,c=−5k
Putting the values of a,b and c in Eq. (i), we get the required equation of plane
⇒5x−5−3y−5z+5=0
⇒‌‌5x−3y−5z=0
Now, plane passing through (11,7,5) and parallel to the plane π(5x−3y−5x).

Equation of a plane parallel to
5x−3y−5z=k......(i)
and passing through (11,7,5)
‌5(11)−3(7)−5(5)‌=k
⇒55−21−25‌=k
⇒k‌=9
So, equation of plane
⇒‌‌5x−3y−5z=9
Comparing with ax+by−z+d=0 (given)
So, a=5,b=−3,c=5d=9