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 planes,

$$2x+ky-5z=1$$

and

$$3kx-ky+z=5$$

These two planes are mutually perpendicular.

Hence, product of normals is zero.

Normal vectors are

$$\vec n_1=(2,k,-5)$$

and

$$\vec n_2=(3k,-k,1)$$

Therefore,

$$\vec n_1\cdot\vec n_2=0$$

$$2(3k)+k(-k)+(-5)(1)=0$$

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

$$k^2-6k+5=0$$

$$(k-1)(k-5)=0$$

Given

$$k<3,$$

hence

$$k=1$$

So the planes become

$$2x+y-5z=1$$

and

$$3x-y+z=5$$

Any plane through their line of intersection is

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

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

The plane intercepts unit length on positive $$x$$-axis.

Hence, putting

$$y=0,\qquad z=0,\qquad x=1$$

must satisfy the plane.

Therefore,

$$2+3\lambda-(1+5\lambda)=0$$

$$1-2\lambda=0$$

$$\lambda=\frac12$$

Substituting,

$$\frac72x+\frac12y-\frac92z-\frac72=0$$

Multiplying by $$2,$$

$$7x+y-9z-7=0$$

Now find the $$y$$-intercept.

Putting

$$x=0,\qquad z=0,$$

$$y-7=0$$

$$y=7$$

Hence, the intercept on the $$y$$-axis is

$$\boxed{7}$$.