If the foot of the perpendicular from the point $A\left(-1,4,3\right)$ on the plane $P:2x+my+nz=4$, is $\left(-2,\frac{7}{2},\frac{3}{2}\right)$, then the distance of the point $A$ from the plane $P$, measured parallel to a line with direction ratios $3,-1,-4$, is equal to

Plotting the data of the given value we have,

Now let $B$ be foot of $\perp$coordinates, so given $B=\left(-2,\frac{7}{2},\frac{3}{2}\right)$

Now direction ratio of line $AB$ will be   

Now comparing with $2x+my+nz=4$ we get, $m=1,n=3$

So equation of $AC$ which is passing through $\left(-1,4,3\right)$ will be

$\frac{x+1}{3}=\frac{y-4}{-1}=\frac{z-3}{-4}=\lambda$

So point $C$ is $\left(3\lambda -1,-\lambda +4,-4\lambda +3\right)$, but given $C$ lies on the plane $2x+y+3z=4$, so on satisfying the point we get,

$6\lambda -2-\lambda +4-12\lambda +9=4$

$\Rightarrow \lambda =1\Rightarrow C\left(2,3,-1\right)$

Now finding the distance between $A \& C$ we get,

$\Rightarrow AC=\sqrt{26}$