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)

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

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

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

\[\therefore 2\left( \frac{5k + 1}{3k + 1} \right) = 3\left( \frac{5k + 1}{4k + 1} \right)\]
\[ \Rightarrow \left( 5k + 1 \right)\left( 8k + 2 - 9k - 3 \right) = 0\]
\[ \Rightarrow \left( 5k + 1 \right)\left( - k - 1 \right) = 0\]
\[ \Rightarrow \left( 5k + 1 \right)\left( k + 1 \right) = 0\]

\[\Rightarrow 5k + 1 = 0 or k + 1 = 0\]
\[ \Rightarrow k = - \frac{1}{5} or k = - 1\]

Putting 

\[k = - \frac{1}{5}\]  in (1), we get

\[\left( - \frac{2}{5} + 1 \right)x + \left( - \frac{3}{5} + 1 \right)y + \left( - \frac{4}{5} + 1 \right)z = 5 \times \left( - \frac{1}{5} \right) + 1\]
\[ \Rightarrow 3x + 2y + z = 0\]

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

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

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.