Three Dimensional Geometry — Class 12 Maths Solution

.: Any plane through the line of intersection of the planes

$x + y + z - 1 = 0$ and $3x + 3y + 4z - 5 = 0$ is

$(x + y + z - 1) + k(2x + 3y + 4z - 5) = 0$
i.e., $(1 + 2k)x + (1 + 3k)y + (1 + 4k)z - (1 + 5k) = 0$ …(1)
Since it is perpendicular to the plane $x - y + z = 0,$ ...(2)
$\therefore$ Their normals are perpendicular.

$\Rightarrow$ $(1 + 2k)(1) + (1 + 3k)( - 1) + (1 + 4k)(1) = 0$

$\Rightarrow$ $1 + 2k - 1 - 3k + 1 + 4k = 0$

$\Rightarrow$ $3k = - 1 \Rightarrow k = - \cfrac{1}{3}$
Putting in (1),

we get
$\left( {1 - \cfrac{2}{3}} \right)x + (1 - 1)y + \left( {1 - \cfrac{4}{3}} \right)z - \left( {1 - \cfrac{5}{3}} \right) = 0$

$\Rightarrow$ $\cfrac{1}{3}x - \cfrac{1}{3}z + \cfrac{2}{3} = 0$

$\Rightarrow$ $x - z + 2 = 0$ ,

which is the required equation.