Sum

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

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#### Solution

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

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

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

\[ \Rightarrow \left( 5k + 1 \right)\left( 8k + 2 - 9k - 3 \right) = 0\]

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

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

\[ \Rightarrow \left( 5k + 1 \right)\left( 8k + 2 - 9k - 3 \right) = 0\]

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

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

\[\Rightarrow 5k + 1 = 0 or k + 1 = 0\]

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

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

Putting

\[k = - \frac{1}{5}\] in (1), we get

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

\[ \Rightarrow 3x + 2y + z = 0\]

\[ \Rightarrow 3x + 2y + z = 0\]

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

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

\[ \Rightarrow - x - 2y - 3z = - 4\]

\[ \Rightarrow x + 2y + 3z = 4\]

\[ \Rightarrow - x - 2y - 3z = - 4\]

\[ \Rightarrow x + 2y + 3z = 4\]

Here, twice of

Thus, the equation of the required plane is

*y*-intercept is equal to three times its*z*-intercept.Thus, the equation of the required plane is

*x*+ 2*y*+ 3*z*= 4. Is there an error in this question or solution?

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