Factorise the following:
$16 x^{2}+4 y^{2}+9 z^{2}-16 x y-12 y z+24 x z$

(N/A) The given expression is $16 x^{2}+4 y^{2}+9 z^{2}-16 x y-12 y z+24 x z$.
We can rewrite this expression using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$.
Here,we observe the terms:
$16 x^{2} = (4 x)^{2}$
$4 y^{2} = (-2 y)^{2}$
$9 z^{2} = (3 z)^{2}$
Now,checking the cross terms:
$2(4 x)(-2 y) = -16 x y$
$2(-2 y)(3 z) = -12 y z$
$2(3 z)(4 x) = 24 x z$
Thus,the expression becomes:
$(4 x)^{2}+(-2 y)^{2}+(3 z)^{2}+2(4 x)(-2 y)+2(-2 y)(3 z)+2(3 z)(4 x)$
Applying the identity,we get:
$(4 x-2 y+3 z)^{2}$
Therefore,the factorised form is $(4 x-2 y+3 z)(4 x-2 y+3 z)$.