Step 1: Find the magnitudes of the vectors
Compute the magnitudes: \[ |\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}, \,\] \[|\vec{b}| = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{35}, \,\]\[ |\vec{c}| = \sqrt{(-3)^2 + 4^2 + 4^2} = \sqrt{41}. \]
Step 2: Check for right angle using dot products
Calculate \( \vec{a} \cdot \vec{b} \), \( \vec{b} \cdot \vec{c} \), and \( \vec{c} \cdot \vec{a} \). If one is zero, the triangle is right-angled. For example: \[ \vec{a} \cdot \vec{c} = (2)(-3) + (-1)(4) + (1)(4) = -6 - 4 + 4 = 0. \]
Step 3: Conclude the type of triangle
Since \( \vec{a} \cdot \vec{c} = 0 \), the triangle is right-angled.
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).