In $\Delta$ GEF and $\Delta$ GCD, we have

∠EFG = ∠GCD (Alternative angle)

∠EFG = ∠CGD(Vertically opposite angles)

$\Delta$ GEF ~ $\Delta$ GCD

Thus,

GE/CG = EF/CD

or, 5/10 = EF/18

or, EF = 9 cm.

Addition any two side of a triangle is always greater then another side then only triangle formation is possible lets e.g a, b, c is a side of a triangle then a+b>c, b+c >a, c+a>b

So option b is not satisfying the condition 3 + 4 $\not\gt $ 8

Orthocentre.

∠ BAL + ∠ B + 90⁰ = 180⁰

or, ∠ BAL + ∠ B = 90⁰

or, ∠ BAL = 90⁰ - ∠ B ----------- (1)

Now, in $\Delta$ ABC,

∠ ACB + ∠ B + ∠ A = 180⁰

∠ ACB = 90⁰

-∠ B ----- (2)

From, (1) and (2),

∠ BAL = ∠ ACB.

∠1 + ∠2 = ∠3 [PS is bisector.] ------ (1)

∠Q = 90⁰ - ∠1

∠R = 90⁰ -∠2 - ∠3

So,

∠Q - ∠R = (90⁰ - ∠1) - (90⁰ - ∠2 - ∠3)

∠Q - ∠R = ∠2 + ∠3 - ∠1

∠Q - ∠ R = ∠2 + (∠1 + ∠2) -∠1[using equation 1]

∠Q - ∠R = 2 ∠2

1/2 * (∠Q - ∠R) = ∠TPS.