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

Given the plane $P$ passes through the intersection of two mutually perpendicular planes $2x+ky-5z=1$ and $3kx-ky+z=5,k<3$

Since planes are mutually perpendicular

Then, $2\left(3k\right)+k\left(-k\right)+\left(-5\right)1=0$

$\Rightarrow 6k-k^2-5=0$

$\Rightarrow k=1,5$

But $k<3$

So $k=1$

Now plane passing through these planes will be,

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

$\Rightarrow \frac{x}{\frac{5\lambda +1}{2+3\lambda }}+\frac{y}{\frac{5\lambda +1}{1-\lambda }}+\frac{z}{\frac{5\lambda +1}{\lambda -5}}=1$

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

So intercept on $y$-axis $=\frac{5\lambda +1}{1-\lambda }=7$