A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then :

<p>$l = {1 \over 2},m = {1 \over {\sqrt 2 }},n = \cos \theta$</p> <p>${l^2} + {m^2} + {n^2} = 1$</p> <p>$$ \Rightarrow {1 \over 4} + {1 \over 2} + {n^2} = 1 \Rightarrow {n^2} = {1 \over 4} \Rightarrow n = \, + \,{1 \over 2}$$</p> <p>$\theta$ is acute $\therefore$ $n = {1 \over 2}$</p> <p>$\therefore$ $$\overrightarrow v = k\left( {{1 \over 2}\widehat i + {1 \over {\sqrt 2 }}\widehat j + {1 \over 2}\widehat k} \right),k \in R$$</p> <p>$\overrightarrow v .\,(\overrightarrow a - \overrightarrow b ) = 0$</p> <p>$$(\sqrt 2 - a){1 \over 2} + ( - 1 - b){1 \over {\sqrt 2 }} + (1 - c){1 \over 2} = 0$$</p> <p>$\Rightarrow {a \over 2} + {b \over {\sqrt 2 }} + {c \over 2} = {1 \over 2}$</p> <p>$\Rightarrow a + \sqrt 2 b + c = 1$</p>