Step 1: Read the parabola.
The parabola is $y^2=12x$. We want the equation of the normal (the line at right angles to the tangent) at a given point.
Step 2: Find $a$.
Compare $y^2=12x$ with the standard form $y^2=4ax$. So $4a=12$, which gives $a=3$.
Step 3: Use the parametric point.
A general point on $y^2=4ax$ is $(at^2,\,2at)$. With $a=3$ this is $(3t^2,\,6t)$. The given point is $(3\lambda^2,\,6\lambda)$, so by matching we get $t=\lambda$.
Step 4: Recall the normal formula.
The standard equation of the normal at parameter $t$ on $y^2=4ax$ is \[ y=-tx+2at+at^3. \] This comes from the fact that the slope of the tangent at $(at^2,2at)$ is $\tfrac{1}{t}$, so the normal slope is $-t$.
Step 5: Substitute the values.
Put $a=3$ and $t=\lambda$: \[ y=-\lambda x+2(3)\lambda+3\lambda^3. \] \[ y=-\lambda x+6\lambda+3\lambda^3. \]
Step 6: Pick the option.
This matches option 1 exactly.
\[ \boxed{\,y=-\lambda x+6\lambda+3\lambda^3\,} \]