(a) Find the foot of the perpendicular drawn from the point ( 1,1,4 ) on the line (x +2)/5=(y +1)/2=(- z +4)/-3.

Let $\displaystyle \frac{\mathrm{x}+2}{5}=\frac{\mathrm{y}+1}{2}=\frac{\mathrm{z}-4}{3}=\lambda$

Coordinate of general point on the given line are $\displaystyle \mathbf{M}(\mathbf{5} \boldsymbol{\lambda}-\mathbf{2}, \mathbf{2} \boldsymbol{\lambda}-\mathbf{1}, \mathbf{3} \boldsymbol{\lambda}+\mathbf{4})$

Direction Ratios of $\displaystyle \mathbf{P M}$ vector are $\displaystyle <\mathbf{5} \boldsymbol{\lambda}-\mathbf{3} \boldsymbol{,} \mathbf{2} \boldsymbol{\lambda} \boldsymbol{-} \mathbf{2} \boldsymbol{,} \mathbf{3} \boldsymbol{\lambda} \boldsymbol{>}$

Since, $\displaystyle \mathbf{P M} \perp \boldsymbol{l}$

$\displaystyle \Rightarrow 5(5 \lambda-3)+2(2 \lambda-2)+3(3 \lambda)=0$

$\displaystyle \Rightarrow \lambda=\frac{1}{2}$

Hence, coordinates of $\displaystyle M$ are $\displaystyle \left(\frac{\mathbf{1}}{\mathbf{2}}, \mathbf{0}, \frac{\mathbf{1 1}}{\mathbf{2}}\right)$

Equation of given line be $\displaystyle \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}+1}{2}=\frac{\mathrm{z}-4}{3}=\boldsymbol{\lambda}$ (say)

Coordinate of any general point on the line are $\displaystyle \mathbf{P}(\mathbf{3} \boldsymbol{\lambda}+\mathbf{1}, \mathbf{2} \boldsymbol{\lambda}-\mathbf{1}, \mathbf{3} \boldsymbol{\lambda}+\mathbf{4})$.

Let distance of point $\displaystyle P$ from ( $\displaystyle -1,-1,2$ ) is $\displaystyle 2 \sqrt{ } 2$.

$\displaystyle \Rightarrow \sqrt{(\mathbf{3} \boldsymbol{\lambda}+\mathbf{2})^{\mathbf{2}}+(\mathbf{2} \boldsymbol{\lambda})^{\mathbf{2}}+(\mathbf{3} \boldsymbol{\lambda}+\mathbf{2})^{\mathbf{2}}}=\mathbf{2} \sqrt{\mathbf{2}}$

$\displaystyle \Rightarrow \mathbf{2 2 \boldsymbol { \lambda } ^ { 2 }}+\mathbf{2 4} \boldsymbol{\lambda}=\mathbf{0}$

$\displaystyle \boldsymbol{\Rightarrow} \boldsymbol{\lambda}=\mathbf{0}$ or $\displaystyle \boldsymbol{\lambda}=-\frac{\mathbf{1 2}}{\mathbf{1 1}}$