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
‌5k(x−1)+(−3k)(y)+(−5k)(z−1)=0
⇒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