"If the plane $P$ passes through the intersection of two mutually perpendicular planes $2 x+k y-5 z=1$ and $3 k x-k y+z=5, k"

Question

Accepted Answer

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${P_1}:2x + ky - 5z = 1$

${P_2}:3kx - ky + z = 5$

$\because$ ${P_1}\, \bot \,{P_2} \Rightarrow 6k - {k^2} + 5 = 0$

$\Rightarrow k = 1,5$

$\because$ $k < 3$

$\therefore$ $k = 1$

${P_1}:2x + y - 5z = 1$

${P_2}:3x - y + z = 5$

$P:(2x + y - 5z - 1) + \lambda (3x - y + z - 5) = 0$

Positive x-axis intercept = 1

$\Rightarrow {{1 + 5\lambda } \over {2 + 3\lambda }} = 1$

$\Rightarrow \lambda = {1 \over 2}$

$\therefore$ $P:7x + y - 4z = 7$

y intercept = 7.

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If the plane $P$ passes through the intersection of two mutually perpendicular planes $2 x+k y-5 z=1$ and $3 k x-k y+z=5, k<3$ and intercepts a unit length on positive $x$-axis, then the intercept made by the plane $P$ on the $y$-axis is :