1. Interpret the condition on $x,y,z$

Since $x,y,z$ are in arithmetic progression with common difference $d$, $y=x+d,\qquad z=x+2d.$

Also given $x=3d$ (the text "x $e$ 3d" clearly means $x=3d$).

Hence, $x=3d,\quad y=4d,\quad z=5d.$

So the determinant becomes

2. Factor out $d$ from the third column

For nonzero $d$, we need

3. Expand the determinant

Using expansion along the first row,

Now compute each minor:

Therefore,

Simplifying,

This becomes $0=0,$ which indicates the determinant is identically zero for all $k$ if we use $x=3d,y=4d,z=5d$.

So let us check the intended interpretation carefully.

4. Likely intended condition: $x\ne 3d$?

The printed statement appears corrupted: "x $e$ 3d". In such problems, the intended relation is usually $x,y,z \text{ are in A.P. with common difference } d,$ so we only use $y=x+d,\quad z=x+2d.$

Then determinant is

Now apply column operation:

Then third column becomes

This is not immediately simplifying nicely.

A better approach is to use the A.P. condition directly as linear dependence in the third column: $2y=x+z.$ So with rows, notice that if determinant is zero, rows are linearly dependent. Let us instead expand in terms of $x,d$ using $y=x+d,\quad z=x+2d.$

5. Compute determinant with $y=x+d, z=x+2d$

Expand along first row:

Now,

So,

Expand:

Group like terms:

• $x$ terms in $\sqrt2$: $15\sqrt2 x+4\sqrt2 x-25\sqrt2 x=-6\sqrt2 x,$
• $d$ terms in $\sqrt2$: $30\sqrt2 d-12\sqrt2 d=18\sqrt2 d,$
• $k$ terms: $-3kx+4kx-3kd=kx-3kd.$

Thus,

Factor:

Hence either $x=3d$ or $k=6\sqrt2.$

Since the given text separately mentions $x=3d$ in corrupted form, the intended conclusion must still be $k=6\sqrt2.$

Therefore, $k^2=(6\sqrt2)^2=72.$

6. Check options

$k^2=72$ which matches Option A.