Three Dimensional Geometry — Class 12 Maths Solution

.: The given planes are $x + y + z - 1 = 0$
and $2x + 3y - z + 4 = 0$

Any plane passing through their intersection is
$(x + y + z - 1) + k(2x + 3y - z + 4) = 0$ ...(1)

Since it is parallel to x-axis.
$\therefore$ Normal to (1) is perpendicular to x-axis.

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

$\Rightarrow$ $1 + 2k = 0 \Rightarrow k = - \cfrac{1}{2}$

Putting in (1),

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

$\Rightarrow$ $2x + 2y + 2x - 2 - 2x - 3y + z - 4 = 0$

$\Rightarrow$ $- y + 3z - 6 = 0$

$\Rightarrow$ $y - 3z + 6 = 0$