Let O be the origin and the position vector of the point P be - ^ i...

Given,

$O$ be the origin and the position vector of the point $P$ be - \overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k},

So, \overset{\to}{OP} = - \overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k}

Also given the position vectors of the points $A, B$ and $C$ are - 2\overset{^}{i} + \overset{^}{j} - 3\overset{^}{k}, 2\overset{^}{i} + 4\overset{^}{j} - 2\overset{^}{k} and - 4\overset{^}{i} ^ + 2\overset{^}{j} - \overset{^}{k}

Now finding,

\overset{\to}{AB} = 4\overset{^}{i} + 3\overset{^}{j} + \overset{^}{k}

\overset{\to}{AC} = - 2\overset{^}{i} + \overset{^}{j} + 2\overset{^}{k}

And \overset{\to}{AB} \times \overset{\to}{AC} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 4 & 3 & 1 \\ - 2 & 1 & 2\end{cases}\right|

⇒ \overset{\to}{AB} \times \overset{\to}{AC} = 5 \overset{^}{i} - 10 \overset{^}{j} + 10 \overset{^}{k}

Now finding the Projection of $\overset{\to}{OP}$ on vector perpendicular to $\overset{\to}{AB} \text{ and } \overset{\to}{AC}$ we get,

Projection$= \left|\dfrac{\overset{\to}{OP} \cdot \left(\overset{\to}{AB} \times \overset{\to}{AC}\right)}{\left|\overset{\to}{AB} \times \overset{\to}{AC}\right|}\right|$

= \dfrac{5\left(- \overset{^}{i} - 2\overset{^}{j} + 3\overset{^}{k}\right)\left(\overset{^}{i} - 2\overset{^}{j} + 2\overset{^}{k}\right)}{5\sqrt{1 + 4 + 4}}$= 3$