The area (in sq.units) of a triangle formed by vertices O, A and B where $\vec{OA} =\hat i+2\hat j+3\hat k$ and $\vec{OB} =-3\hat i-2\hat j+\hat k$ is

$3\sqrt{5}$

The correct answer is Option (1) → $3\sqrt{5}$

Given vectors:

$\vec{OA} = \langle 1,\, 2,\, 3 \rangle$

$\vec{OB} = \langle -3,\,-2,\, 1 \rangle$

Area of triangle = $\frac{1}{2}\,|\vec{OA} \times \vec{OB}|$

Compute cross product:

$\vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{vmatrix}$

$= \hat{i}(2\cdot1 - 3\cdot(-2)) - \hat{j}(1\cdot1 - 3\cdot(-3)) + \hat{k}(1\cdot(-2) - 2\cdot(-3))$

$= \hat{i}(2 + 6) - \hat{j}(1 + 9) + \hat{k}(-2 + 6)$

$= 8\hat{i} - 10\hat{j} + 4\hat{k}$

Magnitude:

$|\vec{OA} \times \vec{OB}| = \sqrt{8^2 + 10^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180} = 6\sqrt{5}$

Area of triangle:

$\text{Area} = \frac{1}{2}(6\sqrt{5}) = 3\sqrt{5}$

The area of the triangle is $3\sqrt{5}$ square units.