Let a plane P contain two lines
$$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$$, $\lambda \in R$ and
$$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$$, $\mu \in R$
If Q($\alpha$, $\beta$, $\gamma$) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3($\alpha$ + $\beta$ + $\gamma$) equals _______.

Given lines,<br><br>$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ parallel to $(\widehat i + \widehat j)$<br><br>Let, $\overrightarrow {{n_1}} = (\widehat i + \widehat j)$<br><br>and $\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$ parallel to $(\widehat j - \widehat k)$<br><br>Let, $\overrightarrow {{n_2}} = (\widehat j - \widehat k)$<br><br>$\therefore$ Normal of plane, $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}}$<br><br>$$\overrightarrow n = \left| {\matrix{ {\widehat i} &amp; {\widehat j} &amp; {\widehat k} \cr 2 &amp; 1 &amp; 0 \cr 0 &amp; 1 &amp; { - 1} \cr } } \right|$$<br><br>$= - \widehat i + \widehat j + \widehat k$<br><br>Line $\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.<br><br>$\therefore$ Equation of the plane, <br><br>$- 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$<br><br>$\Rightarrow x - y - z - 1 = 0$<br><br>Foot of perpendicular from (x₁, y₁, z₁) on the plane,<br><br>$${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$$<br><br>Here foot of perpendicular is drawn from M(1, 0, 1),<br><br>$\therefore$ $${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$$<br><br>$\therefore$ $x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$<br><br>${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$<br><br>${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$<br><br>According to the question,<br><br>$x = \alpha$, $y = \beta$, $z = \gamma$<br><b $$$\beta="-" {1="" \over="" 3}$<br=""><br>$\therefore$ $\alpha = {4 \over 3}$, $\beta = - {1 \over 3}$, $\gamma = {2 \over 3}$<br><br>$\therefore$ $$3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$$</b>