Question

Find the equation of the plane through the line of intersection of the planes $x+y+z=$1 and 2x$+$3$y+$4$z=$5 and twice of its $y$-intercept is equal to three times its $z$-intercept.

Answer 1

The equation of the family of the planes passing through the intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 is

(x + y + z − 1) + k(2x + 3y + 4z − 5) = 0, where k is some constant

⇒ (2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1 .....(1)

$\Rightarrow \frac{x}{\left({\displaystyle \frac{5k+1}{2k+1}}\right)}+\frac{y}{\left({\displaystyle \frac{5k+1}{3k+1}}\right)}+\frac{z}{\left({\displaystyle \frac{5k+1}{4k+1}}\right)}=1$

It is given that twice of y-intercept is equal to three times its z-intercept.

$$\begin{gathered} \therefore 2\left(\frac{5k+1}{3k+1}\right)=3\left(\frac{5k+1}{4k+1}\right) \\ \Rightarrow \left(5k+1\right)\left(8k+2-9k-3\right)=0 \\ \Rightarrow \left(5k+1\right)\left(-k-1\right)=0 \\ \Rightarrow \left(5k+1\right)\left(k+1\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 5k+1=0\mathrm{or}k+1=0 \\ \Rightarrow k=-\frac{1}{5}\mathrm{or}k=-1 \end{gathered}$$
Putting $k=-\frac{1}{5}$ in (1), we get

$$\begin{gathered} \left(-\frac{2}{5}+1\right)x+\left(-\frac{3}{5}+1\right)y+\left(-\frac{4}{5}+1\right)z=5\times \left(-\frac{1}{5}\right)+1 \\ \Rightarrow 3x+2y+z=0 \end{gathered}$$
This plane passes through the origin. So, the intercepts made by the plane with the coordinate axes is 0. Hence, this equation of plane is not accepted as twice of y-intercept is not equal to three times its z-intercept.

Putting $k=-1$ in (1), we get

$$\begin{gathered} \left(-2+1\right)x+\left(-3+1\right)y+\left(-4+1\right)z=5\times \left(-1\right)+1 \\ \Rightarrow -x-2y-3z=-4 \\ \Rightarrow x+2y+3z=4 \end{gathered}$$
Here, twice of y-intercept is equal to three times its z-intercept.

Thus, the equation of the required plane is x + 2y + 3z = 4.