ightarrow \quad & a=\frac{5}{2} \end{aligned}$$

and plane passes through $$(1,1,1)$$.

$$\begin{array}{ll} \therefore & \frac{1}{5 / 2}+\frac{1}{b}+\frac{1}{c}=1 \\ \Rightarrow & \frac{1}{b}+\frac{1}{c}=1-\frac{2}{5} \\ \Rightarrow & \frac{1}{b}+\frac{1}{c}=\frac{3}{5} \quad \text{.... (ii)} \end{array}$$

Perpendicular distance of plane from origin $$=\frac{5}{7}$$

$$\begin{aligned} & \left|\frac{0+0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\right|=\frac{5}{7} \\ & \Rightarrow \quad\left|\frac{-1}{\sqrt{\left(\frac{2}{5}\right)^2+\frac{1}{b^2}+\frac{1}{c^2}}}\right|=\frac{5}{7} \\ & \Rightarrow \quad(-1)^2\left(\frac{7}{5}\right)^2=\left(\frac{2}{5}\right)^2+\frac{1}{b^2}+\frac{1}{c^2} \\ & \Rightarrow \quad \frac{49}{25}-\frac{4}{25}=\frac{1}{b^2}+\frac{1}{c^2} \\ & \Rightarrow \quad \frac{1}{b^2}+\frac{1}{c^2}=\frac{45}{25}=\frac{9}{5} \quad \text{... (iii)}\\ \end{aligned}$$

On squaring Eq. (ii), we get $$ \begin{aligned} & \frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{b c}=\frac{9}{25} \\ & \Rightarrow \quad \frac{9}{5}+\frac{2}{b c}=\frac{9}{25} \quad \text{[from Eq. (iii)]}\\ & \Rightarrow \quad \frac{2}{b c}=\frac{-36}{25} \\ & \Rightarrow b c=-\frac{25}{18} \end{aligned}$$

Putting $$c=-\frac{25}{18} \cdot \frac{1}{b}$$ into Eq. (ii), we get

$$\frac{1}{b}+\frac{1}{\left(\frac{-25}{18} \cdot \frac{1}{b}\right)}=\frac{3}{5}$$

$$\begin{array}{lrl} \Rightarrow & \frac{1}{b}-\frac{18 b}{25} =\frac{3}{5} \\ \Rightarrow & 25-18 b^2 =15 b \\ \Rightarrow & 18 b^2+15 b-25=0 \\ \Rightarrow & 18 b^2+30 b-15 b-25=0 \end{array}$$

$$\begin{array}{lrl} \Rightarrow & 6 b(3 b+5)-5(3 b+5) =0 \\ \Rightarrow & (3 b+5)(6 b-5) =0 \\ \Rightarrow & b =-\frac{5}{3}, \frac{5}{6} \end{array}$$

$$\because y$$-intercept is negative.

$$b=-\frac{5}{3}$$

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