We need to identify:

1. which conic from Column 1 passes through the point of contact $(-1,1)$ for $a=\sqrt2$,
2. which tangent form from Column 2 matches it,
3. which point-expression in Column 3 gives $(-1,1)$.

Then compare with the options.

1. Decode the columns

From the table, the intended entries are:

Column 1

• $(i)$ $x^2+y^2=a$
• $(ii)$ $x^2+a^2y^2=a^2$
• $(iii)$ $y^2=4ax$
• $(iv)$ $x^2-a^2y^2=a^2$

Column 2

• $(i)$ $my=m^2x+a$
• $(ii)$ $y=mx+a\sqrt{m^2+1}$
• $(iii)$ $y=mx+\sqrt{a^2m^2-1}$
• $(iv)$ $y=mx+\sqrt{a^2m^2+1}$

Column 3

Let us label the four point forms as $P,Q,R,S$ in the same order shown:

• $P:\left(\dfrac{a}{m^2},\dfrac{2a}{m}\right)$
• $Q:\left(\dfrac{-ma}{\sqrt{m^2+1}},\dfrac{a}{\sqrt{m^2+1}}\right)$
• $R:\left(\dfrac{-a^2m}{\sqrt{a^2m^2+1}},\dfrac{1}{\sqrt{a^2m^2+1}}\right)$
• $S:\left(\dfrac{-a^2m}{\sqrt{a^2m^2-1}},\dfrac{-1}{\sqrt{a^2m^2-1}}\right)$

From the options, only $P$ and $Q$ are relevant.

2. Put $a=\sqrt2$ and test the point $(-1,1)$ on Column 1 conics

We check each conic.

(i) $x^2+y^2=a$

At $(-1,1)$: $x^2+y^2=1+1=2$ But $a=\sqrt2$, so RHS $=\sqrt2$.

Hence, $2\ne\sqrt2$ So $(-1,1)$ does not lie on conic $(i)$.

(ii) $x^2+a^2y^2=a^2$

Since $a=\sqrt2$, we have $a^2=2$. Thus the conic becomes $x^2+2y^2=2$ At $(-1,1)$: $1+2(1)=3\ne2$ So $(-1,1)$ does not lie on conic $(ii)$.

(iii) $y^2=4ax$

At $(-1,1)$: $1=4\sqrt2(-1)$ which is false. So $(-1,1)$ does not lie on conic $(iii)$.

(iv) $x^2-a^2y^2=a^2$

With $a^2=2$: $x^2-2y^2=2$ At $(-1,1)$: $1-2=-1\ne2$ So $(-1,1)$ does not lie on conic $(iv)$.

This suggests the OCR of Column 1 is distorted. So we must identify the correct intended match using the tangent-point formulas.

3. Use Column 3 point-of-contact formulas

We test which point form can produce $(-1,1)$ when $a=\sqrt2$.

Test $P$

$P=\left(\frac{a}{m^2},\frac{2a}{m}\right)$ For $a=\sqrt2$, $\left(\frac{\sqrt2}{m^2},\frac{2\sqrt2}{m}\right)=(-1,1)$ But first coordinate $\frac{\sqrt2}{m^2}>0$, so it can never be $-1$.

Hence $P$ is impossible.

Test $Q$

$Q=\left(\frac{-ma}{\sqrt{m^2+1}},\frac{a}{\sqrt{m^2+1}}\right)$ Set this equal to $(-1,1)$ with $a=\sqrt2$.

From second coordinate: $\frac{\sqrt2}{\sqrt{m^2+1}}=1$ $m^2+1=2$ $m^2=1$ So $m=\pm1$.

Now first coordinate: $\frac{-m\sqrt2}{\sqrt2}=-m=-1$ Hence $m=1$ So $Q$ indeed gives the point $(-1,1)$.

Therefore Column 3 must be $Q$.

4. Determine corresponding tangent equation

Now use $m=1$ in the tangent forms and see which conic-point pair is standard.

Column 2 item (ii)

$y=mx+a\sqrt{m^2+1}$ For $a=\sqrt2$ and $m=1$: $y=x+\sqrt2\cdot\sqrt2=x+2$ At $(-1,1)$: $1=(-1)+2$ True. So this line passes through the point of contact.

Also, this is the standard tangent to the circle $x^2+y^2=a^2$ with point of contact $\left(\frac{-am}{\sqrt{1+m^2}},\frac{a}{\sqrt{1+m^2}}\right),$ which matches $Q$.

Thus Column 1 must be the circle, i.e. entry $(I)$ as intended in the options.

5. Match with options

We have found:

• Column 1: $(I)$
• Column 2: $(ii)$
• Column 3: $Q$

So the correct combination is: $\boxed{\text{A: }(I),(ii),(Q)}$

6. Compare with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.