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प्रश्न
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
उत्तर
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)
\[ \Rightarrow \left( 5k + 1 \right)\left( 8k + 2 - 9k - 3 \right) = 0\]
\[ \Rightarrow \left( 5k + 1 \right)\left( - k - 1 \right) = 0\]
\[ \Rightarrow \left( 5k + 1 \right)\left( k + 1 \right) = 0\]
\[ \Rightarrow k = - \frac{1}{5} or k = - 1\]
\[ \Rightarrow 3x + 2y + z = 0\]
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
\[ \Rightarrow - x - 2y - 3z = - 4\]
\[ \Rightarrow x + 2y + 3z = 4\]
Thus, the equation of the required plane is x + 2y + 3z = 4.
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