Given lines,

$$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$$ parallel to $$(\widehat i + \widehat j)$$

Let, $$\overrightarrow {{n_1}} = (\widehat i + \widehat j)$$

and $$\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$$ parallel to $$(\widehat j - \widehat k)$$

Let, $$\overrightarrow {{n_2}} = (\widehat j - \widehat k)$$

$$ \therefore $$ Normal of plane, $$\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $$

$$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & 0 \cr 0 & 1 & { - 1} \cr } } \right|$$

$$ = - \widehat i + \widehat j + \widehat k$$

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.

$$ \therefore $$ Equation of the plane,

$$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$$

$$ \Rightarrow x - y - z - 1 = 0$$

Foot of perpendicular from (x₁, y₁, z₁) on the plane,

$${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$$

Here foot of perpendicular is drawn from M(1, 0, 1),

$$ \therefore $$ $${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$$

$$ \therefore $$ $$x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$$

$${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$$

$${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$$

According to the question,

$$x = \alpha $$, $$y = \beta $$, $$z = \gamma $$

$$ \therefore $$ $$\alpha = {4 \over 3}$$, $$\beta = - {1 \over 3}$$, $$\gamma = {2 \over 3}$$

$$ \therefore $$ $$3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$$