Question

Find the equation of the plane which contains the line of intersection of the planes \[x + 2y + 3z - 4 = 0 \text { and } 2x + y - z + 5 = 0\] and whose x-intercept is twice its z-intercept.

Answer 1

Equation of the plane passing through the line of intersection of the given planes is

\[\left( x + 2y + 3z - 4 \right) + \lambda\left( 2x + y - z + 5 \right) = 0\]

\[ \Rightarrow \left( 2\lambda + 1 \right)x + \left( \lambda + 2 \right)y + \left( 3 - \lambda \right)z = 4 - 5\lambda\]

This equation of the plane can be written in the intercept form as

\[\frac{x}{\left( \frac{4 - 5\lambda}{2\lambda + 1} \right)} + \frac{y}{\left( \frac{4 - 5\lambda}{\lambda + 2} \right)} + \frac{z}{\left( \frac{4 - 5\lambda}{3 - \lambda} \right)} = 1\]

It is given that,
x-intercept = 2 × z-intercept

\[\therefore \frac{4 - 5\lambda}{2\lambda + 1} = 2\left( \frac{4 - 5\lambda}{3 - \lambda} \right)\]

\[ \Rightarrow 4\lambda + 2 = 3 - \lambda\]

\[ \Rightarrow \lambda = \frac{1}{5}\]

Therefore, the equation of a plane is 

\[\frac{7}{5}x + \frac{11}{5}y + \frac{14}{5}z = 3\]

\[ \Rightarrow 7x + 11y + 14z = 15\]

Now, the equation of plane parallel to the plane 7x + 11y + 14z = 5 is 7x + 11y + 14z = d.
The plane  7x + 11y + 14z = d passes through (2, 3, −1).
∴ d = 14 + 33 − 14 = 33
Hence, the equation of the required plane is 7x + 11y + 14z = 33.
The vector equation of this plane is `vecr.(7hati + 11hatj + 14hatk)= 33.`