If | |^2=60 and ×( i+2 j+5 k)=overrightarrow0, then a value of ·(-7 i+2 j+3 k) is:

Let, $\vec{c}=a \hat{i}+b \hat{j}+c \hat{k}$c=ai^+bj^​+ck^ Given, $\vec{c} \times(\hat{i}+2 \hat{j}+5 \hat{k})=\overrightarrow{0}$c×(i^+2j^​+5k^)=0 $$\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a & b & c \\ 1 & 2 & 5 \end{array}\right|=\overrightarrow{0} \\ &\Rightarrow(5 b-2 c) \hat{i}-(5 a-c) \hat{j}+(2 a-b) \hat{k} \\ &=0 \hat{i}+0 \hat{j}+0 \hat{k} \end{aligned}$$​⇒​i^a1​j^​b2​k^c5​​=0⇒(5b−2c)i^−(5a−c)j^​+(2a−b)k^=0i^+0j^​+0k^​ Comparing both sides, we get $5 b-2 c=0 ; 5 a-c=0 ; 2 a-b=0$5b−2c=0;5a−c=0;2a−b=0 or $5 b=2 c ; 5 a=c ; 2 a=b$5b=2c;5a=c;2a=b Also given $|\vec{c}|^2=60$∣c∣2=60 $$\Rightarrow a^2+b^2+c^2=60$$⇒a2+b2+c2=60 Putting the value of $b$b and $c$c in above eqn., we get $$\begin{aligned} &a^2+(2 a)^2+(5 a)^2=60 \\ &\Rightarrow a^2+4 a^2+25 a^2=60 \Rightarrow 30 a^2=60 \\ &a^2=2 \\ &a=\pm \sqrt{2} ; b=2 \sqrt{2} ; c=5 \sqrt{2} \end{aligned}$$​a2+(2a)2+(5a)2=60⇒a2+4a2+25a2=60⇒30a2=60a2=2a=±2​;b=22​;c=52​​ Now, $\vec{c}=a \hat{i}+b \hat{j}+c \hat{k}$c=ai^+bj^​+ck^ $$\therefore \vec{c}=\sqrt{2} \hat{i}+2 \sqrt{2} \hat{j}+5 \sqrt{2} \hat{k}$$∴c=2​i^+22​j^​+52​k^ Value of $\vec{c} \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$c⋅(−7i^+2j^​+3k^) is $$\begin{aligned} &(\sqrt{2} \hat{i}+2 \sqrt{2} \hat{j}+5 \sqrt{2} \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\ &=-7 \sqrt{2}+4 \sqrt{2}+15 \sqrt{2}=12 \sqrt{2} \end{aligned}$$​(2​i^+22​j^​+52​k^)⋅(−7i^+2j^​+3k^)=−72​+42​+152​=122​​