The foot of the perpendicular from the point (7,14,5) to the plane...

We know that the length of the perpendicular from the point $\left(x_{1}, y_{1}, z_{1}\right)$ to the plane $a x+b y+c z+d=0$ is $\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$ and the co-ordinate $(\alpha, \beta, \gamma)$ of the foot of the $\perp$ are given by $\frac{\alpha-x_{1}}{a}=\frac{\beta-y_{1}}{b}=\frac{\gamma-z_{1}}{c}=- \left(\frac{a x_{1}+b y_{1}+c z_{1}+d}{a^{2}+b^{2}+c^{2}}\right) ...(1)$ In the given ques, x_{1}=7, y_{1}=14, z_{1}=5$,$a=2 b=4, c=-1, d=-2By putting these values in $(1), we get \frac{\alpha-7}{2}=\frac{\beta-14}{4}=\frac{\gamma-5}{-1}=-\frac{63}{21}\Rightarrow \alpha=1, \beta=2 and$\gamma=$8$ Hence, foot of $\perp$ is (1,2,8)