Find the image of the point having position vector $\hat{i}+3 \hat{j}+4 \hat{k}$ in the plane $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$.

(N/A) Let the given point be $P(\hat{i}+3 \hat{j}+4 \hat{k})$ and $Q$ be the image of $P$ in the plane $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$.
Then $PQ$ is the normal to the plane. Since $PQ$ passes through $P$ and is normal to the given plane,the equation of line $PQ$ is given by $\vec{r}=(\hat{i}+3 \hat{j}+4 \hat{k})+\lambda(2 \hat{i}-\hat{j}+\hat{k})$.
Since $Q$ lies on the line $PQ$,the position vector of $Q$ can be expressed as $(1+2 \lambda) \hat{i}+(3-\lambda) \hat{j}+(4+\lambda) \hat{k}$.
Let $R$ be the point of intersection of the line $PQ$ and the plane. Since $R$ is the midpoint of $PQ$,the position vector of $R$ is $\frac{[(1+2 \lambda) \hat{i}+(3-\lambda) \hat{j}+(4+\lambda) \hat{k}]+[\hat{i}+3 \hat{j}+4 \hat{k}]}{2} = (1+\lambda) \hat{i} + (3-\frac{\lambda}{2}) \hat{j} + (4+\frac{\lambda}{2}) \hat{k}$.
Since $R$ lies on the plane $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$,we have:
$[(1+\lambda) \hat{i} + (3-\frac{\lambda}{2}) \hat{j} + (4+\frac{\lambda}{2}) \hat{k}] \cdot (2 \hat{i}-\hat{j}+\hat{k}) + 3 = 0$
$2(1+\lambda) - (3-\frac{\lambda}{2}) + (4+\frac{\lambda}{2}) + 3 = 0$
$2+2\lambda - 3 + \frac{\lambda}{2} + 4 + \frac{\lambda}{2} + 3 = 0$
$3\lambda + 6 = 0 \Rightarrow \lambda = -2$.
Substituting $\lambda = -2$ into the expression for $Q$:
$Q = (1+2(-2)) \hat{i} + (3-(-2)) \hat{j} + (4+(-2)) \hat{k}$
$Q = -3 \hat{i} + 5 \hat{j} + 2 \hat{k}$.