(a) \( BV = 11.2 \) (M1A1, N2).

Working:
Attempt to use the 3D distance formula (M1)
\( BV = \sqrt{(4 – 8)^2 + (3 – 6)^2 + (10 – 0)^2} \).
\( = \sqrt{(-4)^2 + (-3)^2 + 10^2} = \sqrt{16 + 9 + 100} = \sqrt{125} \).
\( \sqrt{125} = 5\sqrt{5} \approx 11.180 \approx 11.2 \) (3 significant figures) (A1).

[2 marks]

(b) Angle BVC = \( 31.1^\circ \) (M1A1A1A1, N4).

Working:
Attempt to use the cosine rule (M1)
Calculate distances:
\( VC = \sqrt{(8 – 4)^2 + (0 – 3)^2 + (0 – 10)^2} = \sqrt{4^2 + (-3)^2 + (-10)^2} = \sqrt{16 + 9 + 100} = \sqrt{125} \approx 11.180 \).
\( BC = \sqrt{(8 – 8)^2 + (6 – 0)^2 + (0 – 0)^2} = \sqrt{0 + 36 + 0} = 6 \).
Apply cosine rule: \( \cos \theta = \frac{BV^2 + VC^2 – BC^2}{2 \times BV \times VC} \).
\( = \frac{125 + 125 – 36}{2 \times \sqrt{125} \times \sqrt{125}} = \frac{250 – 36}{2 \times 125} = \frac{214}{250} = 0.856 \).
\( \theta = \cos^{-1}(0.856) \approx 31.112^\circ \approx 31.1^\circ \) (3 significant figures) (A1A1A1).

[4 marks]

Total [6 marks]