Question

The plane 4x + 7y + 4z + 81 = 0 is rotated through a right angle about its line of intersection with the plane
5x + 3y + 10z = 25. The equation of the plane in its new position is x – 4y + 6z = k, where k, is

• A. 106
• B. -89
• C. 73
• D. 37

Answer 1

The correct option is A 106

The equation of the plane through the line of intersection of the planes.
4x + 7y + 4z + 81 = 0 and 5x + 3y + 10z = 25
if(4x + 7y + 4z + 81) + λ(5x + 3y + 10z – 25) = 0
$\Rightarrow (4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+81-25\lambda =0....\left(i\right)$
Which is parallel to
x – 4y + 6z = k,
$then\frac{4+5\lambda }{1}=\frac{7+3\lambda }{-4}=\frac{4+10\lambda }{6}weget,\lambda =-1$
Then from Eq. (i),
$-x+4y+6z+106=0$
x – 4y + 6z = 106
Hence, k = 106