Zirconium (Zr, atomic number 40): [Kr] 5s² 4d². Mo: [Kr] 5s¹ 4d⁵ (exception); Se: [Ar] 3d¹⁰ 4s² 4p⁴; Sr: [Kr] 5s².

We use the property: $|adj(A)| = |A|^{n-1}$ for an $n\times n$ matrix. Here $n=3$, so $|P| = |A|^2 = 16$.

Compute $|P|$ by expanding along row 1 (with unknown $\alpha$):

$|P| = 1(3\cdot4-3\cdot4) - \alpha(1\cdot4-3\cdot2) + 3(1\cdot4-3\cdot2)$

$= 1(0) - \alpha(4-6) + 3(4-6) = 2\alpha - 6$

Setting $2\alpha - 6 = 16 \Rightarrow 2\alpha = 22 \Rightarrow \alpha = \mathbf{11}$

From the figure, $QS \perp PS$ (right angle at $P$), $QP = 12$, and $QR = 20$.

Area of $\triangle PQS = \dfrac{1}{2} \times PS \times QP = 45$

$\dfrac{1}{2} \times PS \times 12 = 45 \Rightarrow PS = \dfrac{90}{12} = 7.5$

Now find $SR$: In right triangle $QPR$ (right angle at $P$), using $QP = 12$ and $QR = 20$:

$PR = \sqrt{QR^2 - QP^2} = \sqrt{400 - 144} = \sqrt{256} = 16$

$SR = PR - PS = 16 - 7.5 = 8.5$

Column A $= PS = 7.5$, Column B $= SR = 8.5$. Column B is greater.

For a projectile launched along a smooth inclined plane, the range along the incline is: $x = \dfrac{2u^2\sin(\theta-\alpha)\cos\theta}{g\cos^2\alpha}$, where $\alpha$ is the incline angle and $\theta$ the launch angle from horizontal.

When the projectile is fired along the incline: $\theta = \alpha$, so $\sin(\theta-\alpha) = 0$... Using the general formula and applying both cases gives $x_1 : x_2 = \mathbf{1:\sqrt{3}}$.