Question

Find the equation of the plane passing through the line of intersection of the planes $x+2y+3z-5=0$ and $3x-2y-z+1=0$ and cutting off equal intercepts on the x and z axes.

Answer 1

The equation of any plane passing through the lines of intersection of the given planes

$x+2y+3z-5=0$ and $3x-2y-z+1=0$ is

$(x+2y+3z-5)+k(3x-2y-z+1)=0$ ....... (i)

For the intercept on x-axis, on putting y = 0 and z = 0, we get

$x+3kx-5+k=0\Rightarrow x=\frac{5-k}{3k+1}$

For the intercept on z-axis, on putting x = 0, y = 0, we get

$3z-kz-5+k=0\Rightarrow z=\frac{5-k}{3-k}$

$\therefore$ Intercepts on x-axis and z-axis made by the plane (i) are

$\frac{5-k}{3k+1}$ and $\frac{5-k}{3-k}$ respectively.

Since the intercepts on x-axis and z-axis are equal.

$\frac{5-k}{3k+1}=\frac{5-k}{3-k}$

$\Rightarrow (5-k)(3-k)=(5-k)(3k+1)$

$\Rightarrow 5-k\left[\right(3-k)-(3k+1\left)\right]=0$

$\Rightarrow -(4k-2)(k-5)=0$

$\Rightarrow k=\frac{1}{2},5$

On putting $k=5$ in (i), we notice that the plane passes through origin and hence, it cannot make intercepts on axes. Therefore $k=\frac{1}{2}$ is the only admissible value.

Substituting $k=\frac{1}{2}$ in equation (i) the equation of the required plane is

$x+2y+3z-5+\frac{1}{2}(3x-2y-z+1)=0$

$\Rightarrow 2x+4y+6z-10+3x-2y-z+1=0$

$\Rightarrow 5x+2y+5z-9=0$