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 $5x+3y+10z=25$
$(4x+7y+4z+81)+\lambda (5x+3y+10z-25)=0$
$\Rightarrow (4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+81-25\lambda =0---------------\left(1\right)$
Which is parallel to
$x-4y+6z=k,$
then $\frac{4+5\lambda }{1}=\frac{7+3\lambda }{-4}=\frac{4+10\lambda }{6}$
We get, $\lambda =-1$
Then from Eq. (i),
$-x+4y-6z+106=0$
$\therefore x-4y+6z=106$
Hence, $k=106$